Selina Concise Mathematics Class 10 ICSE Solutions for Chapter 10 - Arithmetic Progression

Exercise 10(A)

1. Which of the following sequences are in arithmetic progression?

i) 2, 6, 10, 14……  

ii) 15, 12, 9, 6…….

iii) 5, 9, 12, 18…… 

iv) $\dfrac{\text{ }\!\! \!\!\text{ 1 }\!\! \!\!\text{ }}{\text{2}}\text{ }\!\! \!\!\text{ ,}\dfrac{\text{ }\!\! \!\!\text{ 1 }\!\! \!\!\text{ }}{\text{3}}\text{ }\!\! \!\!\text{ ,}\dfrac{\text{ }\!\! \!\!\text{ 1 }\!\! \!\!\text{ }}{\text{4}}\text{ }\!\! \!\!\text{ ,}\dfrac{\text{ }\!\! \!\!\text{ 1 }\!\! \!\!\text{ }}{\text{5}}\text{ }\!\! \!\!\text{ }......$   

Ans: 

i) The sequence is 2, 6, 10, 14, .....

Since, 6 - 2 = 4, 10 - 6 = 4 and 14 - 10 = 4.

Difference between the consecutive  terms is the same.

The given sequence is an AP.

The first  term is 12 and the common difference is 14 - 10 = 4.

ii) The sequence is 15, 12, 9, 6 ..... 

Since, 12 - 15 = -3, 9-12 = -3 and 6 - 9 = -3.

Difference between the consecutive  terms is the same.

The given sequence is an AP.

The first  term is 15 and the common difference is 6 - 9 = -3.

iii) The sequence is 5, 9, 12, 18, .....

Since, 9 - 5 = 4, 12 - 9 = 3 and 18 - 12 = 6.

Difference between the consecutive  terms is not the same.

The given sequence is not an AP.

iv) The sequence is $\dfrac{ 1 }{2} ,\dfrac{ 1 }{3} ,\dfrac{ 1 }{4} ,\dfrac{ 1 }{5} ......$

Since, $\dfrac{ 1 }{3}-\dfrac{ 1 }{2}=-\dfrac{ 1 }{6}$, $\dfrac{ 1 }{4}-\dfrac{ 1 }{3}=-\dfrac{ 1 }{12}$, $\dfrac{ 1 }{4}-\dfrac{ 1 }{5}=-\dfrac{ 1 }{20}$

Difference between the consecutive  terms is not the same.

The given sequence is not an AP.

2. The ${{n}^{th}}$  term of a sequence is (2n - 3), find its fifteenth  term.

Ans: The ${{n}^{th}}$  term of a sequence is (2n - 3).

Substitute n = 15,

${{a}_{15}}=\left( 2\times 15-3 \right)$

${{a}_{15}}=\left( 30-3 \right)$

${{a}_{15}}=\left( 27 \right)$

The fifteenth  term of the AP is 27.

3. The ${{p}^{th}}$  term of a sequence is (2p + 3), find the A.P.

Ans: The ${{p}^{th}}$  term of a sequence is (2p + 3).

Substitute n = 1,

${{a}_{1}}=\left( 2\times 1+3 \right)$

${{a}_{1}}=\left( 2+3 \right)$

${{a}_{1}}=\left( 5 \right)$

The first  term of the AP is 5.

Substitute n = 2,

${{a}_{2}}=\left( 2\times 2+3 \right)$

${{a}_{2}}=\left( 4+3 \right)$

${{a}_{2}}=\left( 7 \right)$

The second  term of the AP is 7.

Substitute n = 3,

${{a}_{3}}=\left( 2\times 3+3 \right)$

${{a}_{3}}=\left( 6+3 \right)$

${{a}_{3}}=\left( 9 \right)$

The second  term of the AP is 9.

The A.P. is 5, 7, 9 ……..

4. Find the ${{24}^{th}}$  term of a sequence: 

12, 10, 8, 6……

Ans: The sequence is 12, 10, 8, 6....

The first  term of the sequence is a =12.

The difference between consecutive  terms can be calculated as follows.

${{a}_{2}}-{{a}_{1}}$= 10 - 12 = -2, ${{a}_{3}}-{{a}_{2}}=8-10= -2,  {{a}_{4}}-{{a}_{3}}=6-8= -2$

So the difference between consecutive  terms is the same. So the given sequence is an A.P. and common difference = -2.

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

Substitute 24 for n.

$\Rightarrow {{a}_{24}}=a+\left( 24-1 \right)d$

$\Rightarrow {{a}_{24}}=12+\left( 24-1 \right)\left( -2 \right)$

$\Rightarrow {{a}_{24}}=12-46$

$\Rightarrow {{a}_{24}}= -34$

Hence the ${{24}^{th}}$  term of a sequence is -34.

5. Find ${{30}^{th}}$  term of a sequence:

$\dfrac{ 1 }{2}, 1, \dfrac{3}{ 2 }..........$

Ans: The sequence is $\dfrac{ 1 }{2}, 1, \dfrac{3}{ 2 }..........$

The first  term of the sequence is a =$\dfrac{ 1 }{2}$.

The difference between consecutive  terms can be calculated as follows.

${{a}_{2}}-{{a}_{1}}$= 1 - $\dfrac{ 1 }{2}$ = $\dfrac{ 1 }{2}$ , ${{a}_{3}}-{{a}_{2}}=\dfrac{ 3 }{2}-1= \dfrac{ 1 }{2} $

So the difference between consecutive  terms is the same. So the given sequence is an A.P. and common difference = $\dfrac{ 1 }{2}$.

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

Substitute 30 for n.

$\Rightarrow {{a}_{30}}=a+\left( 30-1 \right)d$

$\Rightarrow {{a}_{30}}=\dfrac{ 1 }{2}+\left( 30-1 \right)\left( \dfrac{ 1 }{2} \right)$

$\Rightarrow {{a}_{30}}=\dfrac{ 1 }{2}+\dfrac{ 29 }{2}$

$\Rightarrow {{a}_{30}}= \dfrac{ 30 }{2}$

$\Rightarrow {{a}_{30}}= 15$

Hence the ${{30}^{th}}$  term of a sequence is 15.

6. Find ${{100}^{th}}$  term of a sequence:

$\sqrt{3}, 2\sqrt{3}, 3\sqrt{3}..........$

Ans: The sequence is $\sqrt{3}, 2\sqrt{3}, 3\sqrt{3}..........$

The first  term of the sequence is a =$\sqrt{3}$.

The difference between consecutive  terms can be calculated as follows.

${{a}_{2}}-{{a}_{1}}$= $2\sqrt{3}$ - $\sqrt{3}$ = $\sqrt{3}$ , ${{a}_{3}}-{{a}_{2}}=3\sqrt{3}-2\sqrt{3}= \sqrt{3} $

So the difference between consecutive  terms is the same. So the given sequence is an A.P. and common difference = $\sqrt{3}$.

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

Substitute 100 for n.

$\Rightarrow {{a}_{100}}=a+\left( 100-1 \right)d$

$\Rightarrow {{a}_{100}}=\sqrt{3} +\left( 100-1 \right)\left( \sqrt{3} \right)$

$\Rightarrow {{a}_{100}}=\sqrt{3}\left( 100 \right)$

$\Rightarrow {{a}_{100}}= 100\sqrt{3}$

Hence the ${{100}^{th}}$ term of a sequence is $ 100\sqrt{3}$.

7. Find ${{50}^{th}}$  term of a sequence:

$\dfrac{1}{ n }, \dfrac{n + 1}{n},\dfrac{2n + 1}{n}..........$

Ans: The sequence is

$\dfrac{1}{ n }, \dfrac{n + 1}{n},\dfrac{2n + 1}{n}..........$

The first term of the sequence is a = $\dfrac{1}{ n }$.

The difference between consecutive terms can be calculated as follows.

${{a}_{2}}-{{a}_{1}}$= $\dfrac{n + 1}{n}$- $\dfrac{1}{ n }$ = 1 , ${{a}_{3}}-{{a}_{2}}=\dfrac{2n + 1}{n}-\dfrac{n + 1}{n}= 1$

So the difference between consecutive terms is the same. So the given sequence is an A.P. and common difference = $1$.

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

Substitute 50 for n.

$\Rightarrow {{a}_{50}}=a+\left( 50-1 \right)d$

$\Rightarrow {{a}_{50}}=\dfrac{1}{ n }+\left( 50-1 \right)\left( 1 \right)$

$\Rightarrow {{a}_{50}}=\dfrac{1}{ n }+49$

$\Rightarrow {{a}_{50}}= \dfrac{49n + 1}{ n }$

Hence the ${{50}^{th}}$  term of a sequence is $\dfrac{49n + 1}{ n }$.

8. Is 402 a  term of the sequence:

8, 13, 18, 23 …….

Ans: The sequence is 8, 13, 18, 23....

The first term of the sequence is a =8.

The difference between consecutive terms can be calculated as follows.

${{a}_{2}}-{{a}_{1}}$= 13 - 8 = 5 , ${{a}_{3}}-{{a}_{2}}=18-13= 5 ,{{a}_{4}}-{{a}_{3}}=23-18= 5$

So the difference between consecutive terms is the same. So the given sequence is an A.P. and common difference = 5.

Now consider 402 is the ${{n}^{th}}$  term of the sequence.

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 402=8+\left( n-1 \right)5$

$\Rightarrow 394=\left( n-1 \right)5$

$\Rightarrow n-1=\dfrac{394}{5}$

$\Rightarrow n=\dfrac{399}{5}$

The number of  terms in any sequence can never be in fraction.

Therefore, 402 is not a  term of the sequence 8, 13, 18, 23,....

9. Find the common difference and ${{99}^{th}}$  term of the Arithmetic progression:

$7\dfrac{3}{4}, 9\dfrac{1}{2}, 11\dfrac{1}{4}...........$

Ans: The sequence is $7\dfrac{3}{4}, 9\dfrac{1}{2}, 11\dfrac{1}{4}...........$

The first  term of the sequence is $7\dfrac{3}{4}$.

The common difference can be obtained as follows,

$\Rightarrow d={{a}_{2}}-{{a}_{1}}=9\dfrac{1}{2}-7\dfrac{3}{4}=\dfrac{19}{2}-\dfrac{31}{4}=\dfrac{38 - 31}{4}=\dfrac{7}{4}$

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

Substitute n = 99

$\Rightarrow {{a}_{99}}=\dfrac{31}{4}+\left( 99-1 \right)\dfrac{7}{4}$

$\Rightarrow {{a}_{99}}=\dfrac{31}{4}+\left( 98 \right)\dfrac{7}{4}$

$\Rightarrow {{a}_{99}}=\dfrac{31}{4}+\dfrac{98 \times  7}{4}$

$\Rightarrow {{a}_{99}}=\dfrac{31 + 98 \times  7}{4}$

$\Rightarrow {{a}_{99}}=\dfrac{31 + 98 \times  7}{4}=179\dfrac{1}{4}$

Hence ${{99}^{th}}$  term of the Arithmetic progression is $179\dfrac{1}{4}$.

10. How many  terms are there in the series:

i) 4, 7, 10, 13, …………, 148?

ii) 0.5, 0.53, 0.56, ………., 1.1?

iii) $\dfrac{3}{4}, 1, 1\dfrac{1}{4}................., 3?$

Ans: 

i) The given series is 4, 7, 10, 13, …………, 148

The first  term of the sequence is 4 and the second  term of the sequence is 7.

The common difference can be calculated as follows.

d = 7 - 4 = 10 - 7

d = 3

Consider the ${{n}^{th}}$  term of the sequence is 148.

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 148=4+\left( n-1 \right)3$

$\Rightarrow 144=\left( n-1 \right)3$

$\Rightarrow \left( n-1 \right)=\dfrac{144}{3}$

$\Rightarrow n=48+1=49$

There are 49  terms in the sequence.

ii) The given series is 0.5, 0.53, 0.56, ………., 1.1

The first term of the sequence is 0.5 and the second  term of the sequence is 0.53.

The common difference can be calculated as follows.

d = 0.53 - 0.5 = 0.56 - 0.53

d = 0.03

Consider the ${{n}^{th}}$  term of the sequence is 1.1

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 1.1=0.5+\left( n-1 \right)\left( 0.03 \right)$

$\Rightarrow 0.6=\left( n-1 \right)\left( 0.03 \right)$

$\Rightarrow \left( n-1 \right)=\dfrac{0.6}{0.03}$

$\Rightarrow n=20+1=21$

There are 21 terms in the sequence.

iii) The given series is $\dfrac{3}{4}, 1, 1\dfrac{1}{4}................., 3$

The first term of the sequence is $\dfrac{3}{4}$and the second  term of the sequence is 1.

The common difference can be calculated as follows.

d = 1 - $\dfrac{3}{4}$ = $\dfrac{5}{4}$ - 1

d = $\dfrac{1}{4}$

Consider the ${{n}^{th}}$ term of the sequence is 3.

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 3=\dfrac{3}{4}+\left( n-1 \right)\left( \dfrac{1}{4} \right)$

$\Rightarrow 3=\dfrac{3}{4}+\dfrac{1}{4}n-\dfrac{1}{4}$

$\Rightarrow \dfrac{n}{4}=3-\dfrac{1}{2}=\dfrac{5}{2}$

$\Rightarrow n=\dfrac{5 \times  4}{2}=10$

There are 10  terms in the sequence.

11. Which term of the A.P. 1, 4, 7, 10, …… is 52?

Ans: The series is 1, 4, 7, 10, ...

The first term of the sequence is 1 and the second  term of the sequence is 4.

The common difference can be calculated as follows,

d = 4 - 1 = 7 - 4

d = 3

Now consider 52 is the ${{n}^{th}}$  term of the A.P.

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 52=1+\left( n-1 \right)3$

$\Rightarrow 51=\left( n-1 \right)3$

$\Rightarrow n-1=17$

$\Rightarrow n=18$

Therefore ${{18}^{th}}$ term of the A.P. is 52.

12. If ${{5}^{th}}$ and ${{6}^{th}}$  terms of an A.P. are respectively 6 and 5, find the ${{11}^{th}}$  term of the A.P.

Ans: The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

The fifth term of the A.P. is 6.

$\Rightarrow 6=a+\left( 5-1 \right)d$

$\Rightarrow 6=a+4d$                     ------(1)

The sixth  term of the A.P. is 5.

$\Rightarrow 5=a+\left( 6-1 \right)d$

$\Rightarrow 5=a+5d$                    -------(2)

Substitute n = 11 in ${{n}^{th}}$  term formula:

$\Rightarrow {{a}_{11}}=a+\left( 11-1 \right)d$

$\Rightarrow {{a}_{11}}=a+\left( 10 \right)d$           -------(3)

Subtract eq(1) from eq(2)-

$\Rightarrow 5-6=a+5d-a-4d$

$\Rightarrow  -1=d$

Substitute d = -1 in eq(2) -

$\Rightarrow 5=a+5\left( -1 \right)$

$\Rightarrow a=10$

Substitute value of a and d in eq(3):

$\Rightarrow {{a}_{11}}=10+\left( 10 \right)\left( -1 \right)$

$\Rightarrow {{a}_{11}}=10-\left( 10 \right)$

$\Rightarrow {{a}_{11}}=0$

The ${{11}^{th}}$  term of the A.P. is 0.

13. If ${{t}_{n}}$represents ${{n}^{th }}$ term of an A.P., ${{t}_{2}}+{{t}_{5}}-{{t}_{3}}=10$ and ${{t}_{2}}$+${{t}_{9}}=17$, find its first  term and common difference.

Ans: Let us assume that the first  term of the AP is ‘a’ and the common difference is ‘d’.

According to the question,

${{t}_{2}}+{{t}_{5}}-{{t}_{3}}=10$

We know that ${{n}^{th }}$ term of an A.P. = $a+\left( n-1 \right)d$

$\Rightarrow a+\left( 2-1 \right)d+a+\left( 5-1 \right)d-a-\left( 3-1 \right)d=10$

$\Rightarrow a+d+2d=10$

$\Rightarrow a+3d=10$----- (1)

${{t}_{2}}$+${{t}_{9}}=17$

$\Rightarrow a+d+a+8d=17$

$\Rightarrow 2a+9d=17$

Substitute value of ‘a’ from eq(1):

$\Rightarrow 2\left( 10-3d \right)+9d=17$

$\Rightarrow 20-6d+9d=17$

$\Rightarrow 3d= -3$

$\Rightarrow d= -1$

Substitute d = -1 in eq(1):

$\Rightarrow a+3\left( -1 \right)=10$

$\Rightarrow a=13$

So, The first term of the AP is 13 and the common difference is -1.

14. Find the ${{10}^{th }}$  term from the end of the A.P.

4, 9, 14, ...... 254.

Ans: Lets reverse the A.P. 254, ……., 14, 9, 4.

Now the first term is 254.

The common difference can be calculated as follows,

d = 9 - 14 = 4 - 9

d = -5

The common difference is 5.

Now we need to calculate the ${{10}^{th }}$ term of reversed A.P.

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

Substitute n = 10

$\Rightarrow {{a}_{10}}=254+\left( 10-1 \right)\left( -5 \right)$

$\Rightarrow {{a}_{10}}=254-45$

$\Rightarrow {{a}_{10}}=209$

Hence the ${{10}^{th }}$  term from the end of the A.P. is 209.

15. Determine the arithmetic progression whose ${{3}^{rd }}$  term is 5 and ${{7}^{th }}$  term is 9.

Ans: The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

Given that, 

The${{3}^{rd }}$  term of AP is 5.

$\Rightarrow 5=a+\left( 3-1 \right)d=a+2d$------(1)

The ${{7}^{th }}$  term of AP is 9.

$\Rightarrow 9=a+6d$--------(2)

Subtract equation (1) from equation (2).

a + 6d -(a + 2d) = 9 - 5

4d = 4

d = 4

d = 1

Substitute 1 for d in equation (1).

a + 2$\times $1 = 5

a + 2 = 5

a = 5 - 2

a = 3 

The first  term of the AP is 3 and the common difference is 1.

The second  term of the AP can be obtained as follows,

${{a}_{2}}=a+d$

$\Rightarrow {{a}_{2}}=3+1=4$

The third term of the AP can be obtained as follows,

${{a}_{3}}=a+2d$

${{a}_{3}}=3+2\left( 1 \right)=5$

So, The arithmetic sequence is 3,4.5,...

16. Find the ${{31}^{st}}$  term of an AP whose ${{10}^{th}}$  term is 38 and ${{16}^{th}}$  term is 74.

Ans: The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

Given that, 

The${{10}^{th}}$  term of AP is 38.

$\Rightarrow 38=a+\left( 10-1 \right)d=a+9d$------(1)

The ${{16}^{th }}$  term of AP is 74.

$\Rightarrow 74=a+15d$--------(2)

Subtract equation (1) from equation (2).

a + 15d - (a + 9d) = 74 - 38

$\Rightarrow $6d = 36

$\Rightarrow $d = 6

Substitute 6 for ‘d’ in equation (1).

a + 9$\times $6 = 38

a = 38 - 54

a = -16

The first  term of the AP is -16 and the common difference is 6.

The ${{31}^{st}}$  term of the AP can be obtained as follows,

${{a}_{31}}=a+30d$

$\Rightarrow {{a}_{31}}= -16+30\left( 6 \right)=180-16$

$\Rightarrow {{a}_{31}}=164$

So, The ${{31}^{st}}$  term of the AP is 164.

17. Which  term of the series: 

21, 18, 15 ……. is -81?

Can any  term of this series be zero? If yes, find the number of  terms.

Ans: The sequence is 21, 18, 15, .....

The first  term of the sequence is 21.

The common difference can be calculated as follows.

d = 18 - 21

d = -3

Consider -81 as ${{n}^{th}}$  term.

$\Rightarrow a+\left( n-1 \right)d= -81$

$\Rightarrow 21+\left( n-1 \right)\left( -3 \right)= -81$

$\Rightarrow \left( n-1 \right)\left( -3 \right)= -81-21$

$\Rightarrow \left( n-1 \right)=\dfrac{-102}{-3}$

$\Rightarrow n=34+1=35$

So, ${{35}^{th}}$  term of the series: 21, 18, 15 ……. is -81.

Consider 0 as ${{n}^{th}}$  term.

$\Rightarrow a+\left( n-1 \right)d= 0$

$\Rightarrow 21+\left( n-1 \right)\left( -3 \right)= 0$

$\Rightarrow \left( n-1 \right)\left( -3 \right)= -21$

$\Rightarrow \left( n-1 \right)=\dfrac{-21}{-3}$

$\Rightarrow n=7+1=8$

So we got the value of n as an integer so we can say that ${{8}^{th}}$ term of AP is zero.

18. An A.P. consists of 60  terms. If the first and last  terms are 7 and 125 respectively, find the ${{31}^{st}}$  term.

Ans: The first  term of the AP is 7.

a = 7

The last/${{60}^{th}}$  term of the AP is 125.

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 125=7+\left( 60-1 \right)d$

$\Rightarrow 118=59d$

$\Rightarrow d=\dfrac{118}{59}=2$

Now the ${{31}^{st}}$  term can be calculated as follows,

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow {{a}_{31}}=7+\left( 31-1 \right)2$

$\Rightarrow {{a}_{31}}=7+60$

$\Rightarrow {{a}_{31}}=67$

So, the ${{31}^{st}}$  term is 67.

19. The sum of the ${{4}^{th}}$ and the ${{8}^{th}}$ terms of an AP is 24 and the sum of the ${{6}^{th}}$and the ${{10}^{th}}$ terms of the same AP is 34. Find the first three  terms of the AP.

Ans: The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

The sum of the fourth  term and eight  terms is 24.

$\Rightarrow {{a}_{4}}+{{a}_{8}}=24$

$\Rightarrow a+3d+a+7d=24$

$\Rightarrow 2a+10d=24$

$\Rightarrow a+5d=12$                          -----(1)

The sum of the sixth  term and tenth  term is 34.

$\Rightarrow {{a}_{6}}+{{a}_{10}}=34$

$\Rightarrow a+5d+a+9d=34$

$\Rightarrow 2a+14d=34$

$\Rightarrow a+7d=17$-----(2)

Subtract eq(1) from eq(2)-

$\Rightarrow a+7d-a-5d=17-12$

$\Rightarrow 2d=5$

$\Rightarrow d=\dfrac{5}{2}$

Substitute value of d in eq(1)-

$\Rightarrow a+5\times \dfrac{5}{2}=12$

$\Rightarrow a=12-\dfrac{25}{2}$

$\Rightarrow a= -\dfrac{1}{2}$

Now we have the first term as $-\dfrac{1}{2}$ and d = $\dfrac{5}{2}$.

Second term can be calculated as follows,

$\Rightarrow {{a}_{2}}=a+d=-\dfrac{1}{2}+\dfrac{5}{2}=2$

Third term can be calculated as follows,

$\Rightarrow {{a}_{3}}=a+2d=-\dfrac{1}{2}+2\times \dfrac{5}{2}=\dfrac{9}{2}=4.5$

So, the first three terms of the AP.

20. If the third term of an A.P. is 5 and the seventh term is 9, find the ${{17}^{th}}$ term.

Ans: The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

Given that, 

The${{3}^{rd }}$  term of AP is 5.

$\Rightarrow 5=a+\left( 3-1 \right)d=a+2d$         --------(1)

The ${{7}^{th }}$  term of AP is 9.

$\Rightarrow 9=a+6d$                                     --------(2)

Subtract equation (1) from equation (2).

a + 6d -(a + 2d) = 9 - 5

4d = 4

d = 4

d = 1

Substitute 1 for d in equation (1).

a + 2$\times $1 = 5

a + 2 = 5

a = 5 - 2

a = 3 

The first  term of the AP is 3 and the common difference is 1.

The ${{17}^{th}}$ term can be calculated as follows,

$\Rightarrow {{a}_{17}}=a+16d$

$\Rightarrow {{a}_{17}}=3+16\times 1$

$\Rightarrow {{a}_{17}}=19$

Chapter-10 - Arithmetic Progression

Exercise 10(B)

1. In an A.P., 10 times of its tenth  term is equal to thirty times of its ${{30}^{th}}$ term. Find the ${{40}^{th}}$ term.

Ans: The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

10 times of its tenth  term is equal to thirty times of its ${{30}^{th}}$ term. 

$\Rightarrow 10\left( {{a}_{10}} \right)=30\left( {{a}_{30}} \right)$

$\Rightarrow 10\left( a+9d \right)=30\left( a+29d \right)$

$\Rightarrow 10a+90a=30a+870d$

$\Rightarrow  -20a=780d$

$\Rightarrow a=\dfrac{  780d }{-20}= -39d$------(1)

Now we can calculate its ${{40}^{th}}$ term as follows,

$\Rightarrow {{a}_{40}}=a+39d$

$\Rightarrow {{a}_{40}}= -39d+39d=0$

So, its ${{40}^{th}}$ term is 0.

2. How many two-digit numbers are divisible by 3?

Ans: The first two digit number that is divisible by 3 is 12.

The last two digit number that is divisible by 3 is 99.

The sequence can be expressed as 12, 15, 18, ..., 99.

The first  term of the A.P is 12.

The common difference can be calculated as follows,

d = 15 -12

d = 3

Consider 99 as the ${{n}^{th}} $ term.

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 99=12+\left( n-1 \right)3$

$\Rightarrow \dfrac{87}{3}=n-1$

$\Rightarrow n=30$

There are 30  terms. So there are 30 two-digit numbers divisible by 3.

3. Which term of an A.P. 5, 15, 25, ……. Will it be 130 more than its ${{31}^{st}}$ term?

Ans: The A.P is 5, 15, 25, .....

The first term is 5.

The common difference can be calculated as follows,

d = 15 - 5

d =10

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

Consider the term that is 130 more than its ${{31}^{st}}$  term is ${{n}^{th}} $ term.

$\Rightarrow a+\left( n-1 \right)d=a+30d+130$

$\Rightarrow \left( n-1-30 \right)10=130$

$\Rightarrow \left( n-31 \right)=13$

$\Rightarrow n=44$

So ${{44}^{th}}$ term of an A.P. 5, 15, 25, ……. Will be 130 more than its ${{31}^{st}}$ term.

4. Find the value of p, if x and 2x + p and 3x + 6 are in A.P.

Ans: $x, 2x+p,3x+6$ are in A.P.

The common differences between the terms are equal.

$\Rightarrow 2x+p-x=3x+6-2x-p$

$\Rightarrow p+p=6$

$\Rightarrow p=3$

So, the value p is 3.

5. If the ${{3}^{rd}}$ and ${{9}^{th}} $ terms of an arithmetic progression are 4 and -8 respectively, which  term of it is zero?

Ans: The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

So, 

The ${{3}^{rd}}$ term of AP is 4.

$\Rightarrow a+2d=4$                        ------(1)

The ${{9}^{th}}$ term of AP is -8.

$\Rightarrow a+8d=-8$                    ------(2)

Substituting value of a from eq(1)-

$\Rightarrow 4-2d+8d= -8$

$\Rightarrow 6d= -12$

$\Rightarrow d= -2$

Substituting value of d in eq(1)-

$\Rightarrow a+2\left( -2 \right)=4$

$\Rightarrow a=8$

Now, let ${{n}^{th}}$  term of AP is zero.

$\Rightarrow a+\left( n-1 \right)d=0$

$\Rightarrow 8+\left( n-1 \right)\left( -2 \right)=0$

$\Rightarrow n-1=4$

$\Rightarrow n=5$

So the fifth term of AP is zero.

6. How many three digit numbers are divisible by 87?

Ans: The first three digit number that is divisible by 87 is 174.

The last three digit number that is divisible by 87 is 957.

The sequence can be expressed as 174,261,348,...,957.

The first  term of the A.P is 174.

The common difference can be calculated as follows,

d = 261 -174

d = 87

Consider 957 as the ${{n}^{th}} $ term.

The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 957=174+\left( n-1 \right)87$

$\Rightarrow \dfrac{783}{87}=n-1$

$\Rightarrow n=9+1=10$

There are 10 terms. So there are 10 three-digit numbers divisible by 87.

7. For what value of n, the ${{n}^{th}}$ term of the AP 63, 65, 67……. and the ${{n}^{th}}$ term of the AP 3, 10, 17……. are equal to each other?

Ans: Consider the first term of the sequence 63, 65, 67.... as A and

consider the common difference as D.

First term of the A.P is A = 63.

The common difference can be calculated as follows,

D = 65 - 63 = 2

Consider the first term of the sequence 3, 10, 17,... as a and consider the common difference as d.

First term of the A.P is a = 3.

The common difference can be calculated as follows,

d = 10 - 3 = 7

As given ${{n}^{th }}$ terms of both APs are equal.

$\Rightarrow {{A}_{n}}={{a}_{n}}$

$\Rightarrow A+\left( n-1 \right)D=a+\left( n-1 \right)d$

$\Rightarrow 63+\left( n-1 \right)2=3+\left( n-1 \right)7$

$\Rightarrow 60=\left( n-1 \right)5$

$\Rightarrow n-1=\dfrac{60}{5}$

$\Rightarrow n=12+1=13$

So the value of n = 13, the ${{n}^{th}}$ term of the AP 63, 65, 67……. and the ${{n}^{th}}$ term of the AP 3, 10, 17……. are equal to each other.

8. Determine the AP whose ${{3}^{rd}}$  term is 16 and ${{7}^{th}}$  term exceeds ${{5}^{th}}$  term by 12.

Ans: The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

The third  term of an AP is 16.

$\Rightarrow {{a}_{3}}=16$

$\Rightarrow a+2d=16$                      ------(1)

The${{7}^{th}}$  term exceeds ${{5}^{th}}$  term by 12.

$\Rightarrow {{a}_{7}}={{a}_{5}}+12$

$\Rightarrow a+6d=a+4d+12$

$\Rightarrow 2d=12$

$\Rightarrow d=6$

Substitute d = 6 in eq(1)-

$\Rightarrow a+2\left( 6 \right)=16$

$\Rightarrow a=4$

So from the values of ‘a’ and ‘d’ we will get the  terms of AP.

Second term:

${{a}_{2}}=a+d=4+6=10$

So, AP is 4, 10, 16, ………

9. If numbers n - 2, 4n - 1 and 5n+2 are in AP, find the value of n and its next two terms.

Ans: n - 2, 4n - 1, 5n+2 are in A.P.

The common differences between the  terms are equal.

$\Rightarrow 4n-1-\left( n-2 \right)=5n+2-\left( 4n-1 \right)$

$\Rightarrow 3n+1=n+3$

$\Rightarrow 2n=2$

$\Rightarrow n=1$

So, first term = n - 2 = 1 - 2 = -1

Second term = 4n - 1= 4 - 1 = 3

Third term = 5n+2 = 5+2 = 7

Common difference = 7 - 3 = 4

Fourth term = Third term + Common difference

Fourth term = 7+4 = 11

Fifth term = Fourth term + Common difference

Fifth term = 11+4 = 15

10. Determine the value of k for which ${{k}^{2}}+4k+8, 2{{k}^{2}}+3k+6$and$3{{k}^{2}}+4k+4$are in AP.

Ans: ${{k}^{2}}+4k+8, 2{{k}^{2}}+3k+6$and$3{{k}^{2}}+4k+4$are in AP.

The common differences between the terms are equal.

$2{{k}^{2}}+3k+6-({{k}^{2}}+4k+8)=3{{k}^{2}}+4k+4-\left( 2{{k}^{2}}+3k+6 \right)$

$\Rightarrow {{k}^{2}}-k-2={{k}^{2}}+k-2$

$\Rightarrow 0=2k$

$\Rightarrow k=0$

The value of k is 0.

11. If a, b and c are in AP then show that:

i) 4a, 4b and 4c are in AP.

ii) a+4, b+4 and c+4 are in AP.

Ans: If a, b and c are in AP then the common differences between the  terms are equal.

$\Rightarrow b-a=c-b$

$\Rightarrow 2b=a+c$             ………(1)

i) For the  terms 4a,4b and 4c to be in A.P the common difference will be equal.

$\Rightarrow 4b - 4a=4c-4b$

$\Rightarrow 4b + 4b=4c+4a$

$\Rightarrow 8b=4c+4a$

$\Rightarrow 2b=a+c$

So it is eq(1) so the  terms 4a,4b and 4c are in A.P.

ii) For the  terms a+4, b+4 and c+4 to be in A.P the common difference will be equal.

$\Rightarrow b+4 -\left( a+4 \right)=c+4-\left( b+4 \right)$

$\Rightarrow b-a=c-b$

$\Rightarrow 2b=a+c$

So it is eq(1) so the  terms a+4, b+4 and c+4 are in A.P.

12. An AP consists of 57  terms of which ${{7}^{th}}$ term is 13 and the last  term is 108. Find the ${{45}^{th}}$  term of this AP.

Ans: The seventh  term of an AP is 13.

$\Rightarrow {{a}_{7}}=13$

$\Rightarrow a+6d=13$           ……..(1)

The last  term of this AP is 108.

$\Rightarrow {{a}_{57}}=108$

$\Rightarrow a+56d=108$            ………(2)

Substituting value of ‘a’ from eq(1) in eq(2)-

$\Rightarrow 13-6d+56d=108$

$\Rightarrow 50d=95$

$\Rightarrow d=\dfrac{95}{50}=1.9$

Substituting value of ‘d’ in eq(1)-

$\Rightarrow a+6\left( 1.9 \right)=13$

$\Rightarrow a=13-11.4$

$\Rightarrow a=1.6$

The ${{45}^{th}}$ term can be calculated as follows,

${{a}_{45}}=a+44d$

$\Rightarrow {{a}_{45}}=1.6+44\left( 1.9 \right)$

$\Rightarrow {{a}_{45}}=1.6+83.6$

$\Rightarrow {{a}_{45}}=85.2$

13. ${{4}^{th}}$  term of an AP is equal to 3 times its first  term and ${{7}^{th}}$ term exceeds twice the ${{3}^{rd}}$ term by 1. Find the first  term and common difference.

Ans: Let the first  term of AP is ‘a’ and the common difference is ‘d’.

The${{4}^{th}}$ term of an AP is equal to 3 times its first  term.

$\Rightarrow {{a}_{4}}=3a$

$\Rightarrow a+3d=3a$

$\Rightarrow 3d=2a$

$\Rightarrow \dfrac{3}{2}d=a$           …………..(1)

The${{7}^{th}}$ term exceeds twice the ${{3}^{rd}}$ term by 1.

$\Rightarrow {{a}_{7}}=2{{a}_{3}}+1$

$\Rightarrow a+6d=2\left( a+2d \right)+1$

$\Rightarrow a-2a+6d-4d=1$

$\Rightarrow  -a+2d=1$

Substituting value of a from eq(1)-

$\Rightarrow  -\dfrac{3}{2}d+2d=1$

$\Rightarrow \dfrac{d}{2}=1$

$\Rightarrow d=2$

From eq(1)-

$\Rightarrow a=\dfrac{3}{2}\times 2=3$

The first  term and common difference are 3 and 2 respectively.

14. The sum of the ${{2}^{nd}}$ and the ${{7}^{th}}$  term of an AP is 30. If it's ${{15}^{th}}$  term is 1 less than twice of its ${{8}^{th}}$ term, find the AP.

Ans: The sum of the ${{2}^{nd}}$ and the${{7}^{th}}$ term of an AP is 30.

$\Rightarrow {{a}_{2}}+{{a}_{7}}=30$

$\Rightarrow a+d+a+6d=30$

$\Rightarrow 2a+7d=30$              ……..(1)

It's ${{15}^{th}}$  term is 1 less than twice its ${{8}^{th}}$ term.

$\Rightarrow {{a}_{15}}=2{{a}_{8}}-1$

$\Rightarrow a+14d=2\left( a+7d \right)-1$

$\Rightarrow a+14d=2a+14d-1$

$\Rightarrow a=2a-1$

$\Rightarrow a=1$

Substituting $a=1$ in eq(1)-

$\Rightarrow 2\left( 1 \right)+7d=30$

$\Rightarrow 7d=28$

$\Rightarrow d=4$

Now second term is calculated as follows,

$\Rightarrow {{a}_{2}}=a+d$

$\Rightarrow {{a}_{2}}=1+4=5$

Now third term is calculated as follows,

$\Rightarrow {{a}_{3}}=a+2d$

$\Rightarrow {{a}_{3}}=1+2\left( 4 \right)=9$

So, the AP is 1, 5, 9 ………

15. In an AP, if ${{m}^{th}}$  term is ‘n’ and ${{n}^{th}}$  term is ‘m’, show that its ${{r}^{th}}$ term is (m+n-r).

Ans: The ${{m}^{th}}$  term is ‘n’.

$\Rightarrow a+\left( m-1 \right)d=n$

$\Rightarrow a+md-d=n$       ………(1)

The ${{n}^{th}}$  term is ‘m’.

$\Rightarrow a+\left( n-1 \right)d=m$

$\Rightarrow a+nd-d=m$         ………(2)

Subtract eq(1) from eq(2)-

$\Rightarrow a+md-d-\left( a+nd-d \right)=n-m$

$\Rightarrow \left( m-n \right)d= -\left( m-n \right)$

$\Rightarrow d= -1$

Substitute $d= -1$in eq(1)-

$\Rightarrow a+m\left( -1 \right)-\left( -1 \right)=n$

$\Rightarrow a=m+n-1$

Now ${{r}^{th}}$  term can be calculated as follows, 

${{a}_{r}}=a+\left( r-1 \right)d$

$\Rightarrow {{a}_{r}}=a+\left( r-1 \right)\left( -1 \right)$

$\Rightarrow {{a}_{r}}=m+n-1-\left( r-1 \right)$

$\Rightarrow {{a}_{r}}=m+n-r$

Hence proved.

16. Which term of the AP 3, 10, 17….... will be 84 more than its ${{13}^{th}}$  term?

Ans: The given AP 3, 10, 17…...

The first term of given AP = 3

Common difference = 10 - 3 = 7

Let ${{n}^{th}}$  term be 84 more than its ${{13}^{th}}$  term.

$\Rightarrow {{a}_{n}}={{a}_{13}}+84$

$\Rightarrow a+\left( n-1 \right)d=a+12d+84$

$\Rightarrow \left( n-1-12 \right)7=84$

$\Rightarrow \left( n-13 \right)=\dfrac{84}{7}$

$\Rightarrow \left( n-13 \right)=12$

$\Rightarrow n=25$

So ${{25}^{th}}$  term of the AP 3, 10, 17….... will be 84 more than its ${{13}^{th}}$  term.

Exercise 10(C)

1. Find the sum of the first 22  terms of the AP: 8, 3, -2 ……

Ans: The AP is 8, 3, -2 ……

The first term of the A.P is 8.

The second term of the A.P is 3.

The common difference can be calculated as follows,

d = 3 - 8

d = -5

The sum of first n term of the A.P can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

The sum of first 22  term of the A.P can be calculated as follows,

$\Rightarrow {{S}_{22}}=\dfrac{22}{2}\left[ 2\left( 8 \right)+\left( 22-1 \right)\left( -5 \right) \right]$

$\Rightarrow {{S}_{22}}=11\left[ 16+\left( 21 \right)\left( -5 \right) \right]$

$\Rightarrow {{S}_{22}}=11\left[ 16-105 \right]$

$\Rightarrow {{S}_{22}}=11\left[ -89 \right]$

$\Rightarrow {{S}_{22}}= -979$

So, the sum of the first 22  terms of the AP: 8, 3, -2 …… is -979.

2. How many terms of the AP: 24, 21, 18 …. must be taken so that their sum is 78?

Ans: AP: 24, 21, 18 …. 

The first  term of the A.P is 24.

The second  term of the A.P is 21.

The common difference can be calculated as follows,

d = 21 - 24

d = -3

Let the sum of the first n  terms is 78.

The sum of first n term of the A.P can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]=78$

$\Rightarrow \dfrac{n}{2}\left[ 2\left( 24 \right)+\left( n-1 \right)\left( -3 \right) \right]=78$

$\Rightarrow n\left[ 48-3n+3 \right]=156$

$\Rightarrow n\left[ 51-3n \right]=156$

$\Rightarrow 3{{n}^{2}}-51n+156=0$

Solve the quadratic equation for the value of n.

$\Rightarrow 3{{n}^{2}}-51n+156=0$

$\Rightarrow {{n}^{2}}-17n+52=0$

$\Rightarrow {{n}^{2}}-13n-4n+52=0$

$\Rightarrow n\left( n-13 \right)-4\left( n-13 \right)=0$

$\Rightarrow \left( n-13 \right)\left( n-4 \right)=0$

$\Rightarrow n=4\And 13$

So either 4 or 13 terms must be taken to get the sum of  terms of the AP is 78.

3. Find the sum of 28 terms of an AP whose ${{n}^{th}}$ term is 8n - 5.

Ans: The${{n}^{th}}$ term of AP = 8n - 5.

First  term:

Substitute n = 1

$\Rightarrow {{a}_{1}}=a=8-5=3$

Second  term:

Substitute n = 2

$\Rightarrow {{a}_{2}}=8\left( 2 \right)-5=11$

Common difference = ${{a}_{2}}-{{a}_{1}}=11-3=8$

The sum of first n  term of the A.P can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

Substitute n = 28

$\Rightarrow {{S}_{28}}=\dfrac{28}{2}\left[ 2\left( 3 \right)+\left( 28-1 \right)8 \right]$

$\Rightarrow {{S}_{28}}=14\left[ 6+\left( 27 \right)8 \right]$

$\Rightarrow {{S}_{28}}=14\left[ 222 \right]$

$\Rightarrow {{S}_{28}}=3108$

The sum of 28  terms of an AP is 3108.

4. Find the sum of:

i) all odd natural numbers less than 50.

ii) first 12 natural numbers each of which is a multiple of 7.

Ans: i) The sequence of odd number less than 50 can be expressed as,

1, 3, 5, … 49.

There are 25 terms in the A.P. 

The sum of odd number less than 50 can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2}\left[ a+{{a}_{n}} \right]$

$\Rightarrow {{S}_{25}}=\dfrac{25}{2}\left[ 1+49 \right]$

$\Rightarrow {{S}_{25}}=\dfrac{25}{2}\left[ 50 \right]$

$\Rightarrow {{S}_{25}}=25\times 25$

$\Rightarrow {{S}_{25}}=625$

ii) The first 12 natural numbers each of which is a multiple of 7.

The sequence can be expressed as follows,

7, 14, 21, ... 84

The sum of first 12 natural numbers each of which is a multiple of 7 can be obtained as follows,

${{S}_{n}}=\dfrac{n}{2}\left[ a+{{a}_{n}} \right]$

$\Rightarrow {{S}_{12}}=\dfrac{12}{2}\left[ 7+84 \right]$

$\Rightarrow {{S}_{12}}=6\left[ 91 \right]$

$\Rightarrow {{S}_{L12}}=546$

5. Find the sum of the first 51  terms of an AP whose ${{2}^{nd }}$and ${{3}^{rd }}$ terms are 14 and 18 respectively.

Ans: The formula for ${{n}^{th}}$  term of the A.P is given as follows.

${{a}_{n}}=a+\left( n-1 \right)d$

The second  term of AP is 14.

$\Rightarrow {{a}_{2}}=a+d=14$                      ………….(1)

The third  term of AP is 18.

$\Rightarrow {{a}_{3}}=a+2d=18$                   …………(2)

Subtract eq(1) from eq(2)-

$a+2d-a-d=18-14$

$\Rightarrow d=4$

From eq(1)-

$\Rightarrow a+4=14$

$\Rightarrow a=10$

The sum of first n  term of the A.P can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

Substitute the values in formula-

$\Rightarrow {{S}_{51}}=\dfrac{51}{2}\left[ 2\left( 10 \right)+\left( 51-1 \right)4 \right]$

$\Rightarrow {{S}_{51}}=\dfrac{51}{2}\left[ 20+200 \right]$

$\Rightarrow {{S}_{51}}=\dfrac{51}{2}\left[ 220 \right]$

$\Rightarrow {{S}_{51}}=51\times 110$

$\Rightarrow {{S}_{51}}=5610$

6. The sum of the first 7 terms of an AP is 49 and that of the first 17  terms of it is 289. Find sum of first n terms.

Ans: The sum of first n  term of the A.P can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

Substitute n = 7,

${{S}_{7}}=\dfrac{7}{2}\left[ 2a+\left( 7-1 \right)d \right]$

$\Rightarrow 49=\dfrac{7}{2}\left[ 2a+6d \right]$

$\Rightarrow 7=a+3d$                       ………….(1)

Substitute n = 17,

${{S}_{17}}=\dfrac{17}{2}\left[ 2a+\left( 17-1 \right)d \right]$

$\Rightarrow 289=\dfrac{17}{2}\left[ 2a+16d \right]$

$\Rightarrow 17=a+8d$                      ……………..(2)

Subtract eq (1) from eq (2)-

$\Rightarrow 17-7=a+8d-a-3d$

$\Rightarrow 10=5d$

$\Rightarrow d=2$

From eq (1)-

$\Rightarrow 7=a+3\left( 2 \right)$

$\Rightarrow a=7-6=1$

So, the sum of first n  terms:

${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2\left( 1 \right)+\left( n-1 \right)2 \right]$

$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2+2n-2 \right]$

$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2n \right]$

So the sum of n term $:{{S}_{n}}={{n}^{2}}$

7. The first term of an AP is 5, the last  term is 45 and the sum of its  terms is 1000. Find the number of terms and the common difference of the AP.

Ans: The first term of an AP is a = 5

the last term of an AP is $l$ = 45

Let the number of terms in AP is n.

The sum of n term ${{S}_{n}}=\dfrac{n}{2} \left[  a+l  \right]$

The sum of n term ${{S}_{n}}=1000$

$\Rightarrow 1000=\dfrac{n}{2}\times \left[ 5+45 \right]$

$\Rightarrow 2000=50n$

$\Rightarrow n=\dfrac{2000}{50}=40$

So the number of terms in AP is 40.

So last term/ ${{40}^{th}}$ term is 45.

$\Rightarrow a+\left( 40-1 \right)d=45$

$\Rightarrow 39d=45-5$

$\Rightarrow d=\dfrac{40}{39}$

So the common difference of the AP is $\dfrac{40}{39}$.

8. Find the sum of all natural numbers between 250 and 1000 which are divisible by Ans: The first number which is greater than 250 and divisible by 9 = 252

The number which is just less than 1000 and divisible by 9 = 999

The sequence can be expressed as follows,

252, 261, 270 …… 999

The number of terms of the AP can be calculated as follows,

$\Rightarrow {{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 999=252+\left( n-1 \right)9$

$\Rightarrow 999-252=\left( n-1 \right)9$

$\Rightarrow \dfrac{747}{9}=\left( n-1 \right)$

$\Rightarrow n=83+1=84$

The sum of first n  term of the A.P can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

$\Rightarrow {{S}_{84}}=\dfrac{84}{2}\left[ 2\left( 252 \right)+\left( 84-1 \right)9 \right]$

$\Rightarrow {{S}_{84}}=42\left[ 504+\left( 83 \right)9 \right]$

$\Rightarrow {{S}_{84}}=42\left[ 504+747 \right]$

$\Rightarrow {{S}_{84}}=42\left[ 1251 \right]$

$\Rightarrow {{S}_{84}}=52,542$

So, The sum of  terms is 52542.

9. The first and last term of an AP are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?

Ans: First term of an A.P. is 34.

Common difference is 18.

Let 700 be the ${{n}^{th}}$  term.

$\Rightarrow {{a}_{n}}=700$

$\Rightarrow a+\left( n-1 \right)d=700$

$\Rightarrow 34+\left( n-1 \right)18=700$

$\Rightarrow n-1=\dfrac{700-34}{18}$

$\Rightarrow n=37+1=38$

The sum of n terms of an AP can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2} \left[  a+{{a}_{n}}  \right]$

$\Rightarrow {{S}_{n}}=\dfrac{038}{2} \left[  34+700  \right]$

$\Rightarrow {{S}_{n}}=19 \left[  734  \right]$

$\Rightarrow {{S}_{n}}=13,946$

10. In an AP, the first  term is 25 and ${{n}^{th}}$ term is -17 and the sum of n  terms is 132. Find n and the common difference.

Ans: The first term of AP is 25.

The ${{n}^{th}}$ term of AP is -17.

The sum of n terms is 132.

The sum of n terms of an AP can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2} \left[  a+{{a}_{n}}  \right]$

$\Rightarrow 132=\dfrac{n}{2} \left[ 25+\left( -17 \right) \right]$

$\Rightarrow 264=n\left[ 8 \right]$

$\Rightarrow n=\dfrac{264}{8}=33$

The ${{n}^{th }}$ term can be calculated as follows,

$\Rightarrow {{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow  -17=25+\left( 33-1 \right)d$

$\Rightarrow  -42=32d$

$\Rightarrow d= -\dfrac{21}{16}$

11. If the ${{8}^{th}}$ term of AP is 37 and the ${{15}^{th}}$ term is 15 more than the ${{12}^{th}}$ term, find the AP. Also find the sum of the first 20  terms of this AP.

Ans:  The ${{8}^{th}}$ term of AP is 37.

$\Rightarrow {{a}_{8}}=37$

$\Rightarrow a+7d=37$                 ………….(1)

The ${{15}^{th}}$ term is 15 more than the${{12}^{th}}$ term.

$\Rightarrow {{a}_{15}}={{a}_{12}}+15$

$\Rightarrow a+14d=a +11d+15$

$\Rightarrow 3d=15$

$\Rightarrow d=5$

From eq(1)-

$\Rightarrow a+7\left( 5 \right)=37$

$\Rightarrow a=37-35=2$

Second term:

$\Rightarrow {{a}_{2}}=a+d$

$\Rightarrow {{a}_{2}}=2+5=7$

Third term:

$\Rightarrow {{a}_{3}}=a+2d$

$\Rightarrow {{a}_{3}}=2+2\left( 5 \right)=12$

So, the AP is 2, 7, 12, ……....

The sum of n terms of an AP can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2} \left[  2a+\left( n-1 \right)d  \right]$

$\Rightarrow {{S}_{20}}=\dfrac{20}{2} \left[  2\times 2+\left( 20-1 \right)5  \right]$

$\Rightarrow {{S}_{20}}=10 \left[  4+95  \right]$

$\Rightarrow {{S}_{20}}=990$

So, the sum of the first 20 terms of this AP is 990.

12. Find the sum of all multiples of 7 lying between 300 and 700.

Ans: The first number which is greater than 300 and multiple of 7 = 301

The number which is just less than 700 and multiple of 7 = 693

The sequence can be expressed as follows,

301, 308, 315 …… 693

The number of terms of the AP can be calculated as follows,

$\Rightarrow {{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 693=301+\left( n-1 \right)7$

$\Rightarrow 693-301=\left( n-1 \right)7$

$\Rightarrow \dfrac{ 392 }{7 }=\left( n-1 \right)$

$\Rightarrow n=56+1=57$

The sum of first n term of the A.P can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2}\left[ a+{{a}_{n}} \right]$

$\Rightarrow {{S}_{57}}=\dfrac{57}{2}\left[ 301+693 \right]$

$\Rightarrow {{S}_{57}}=\dfrac{57}{2}\left[ 994 \right]$

$\Rightarrow {{S}_{84}}=57\times 497$

$\Rightarrow {{S}_{84}}=28329$

So, the sum of terms is 28329.

13 Sum of n natural numbers is $5{{n}^{2}}+4n$. Find its ${{8}^{th}}$ term.

Ans:  Sum of n natural numbers is $5{{n}^{2}}+4n$.

${{8}^{th}}$ term can be calculated as follows,

${{a}_{8}}={{S}_{8}}-{{S}_{7}}$

$\Rightarrow {{a}_{8}}=5{{(8)}^{2}}+4\left( 8 \right)-[5\left( 7{{)}^{2}}+4\left( 7 \right) \right]$

$\Rightarrow {{a}_{8}}=5[{{(8)}^{2}}-\left( 7{{)}^{2}} \right]+4\left( 8-7 \right)$

$\Rightarrow {{a}_{8}}=5\left[ 64-49 \right]+4\left( 1 \right)$

$\Rightarrow {{a}_{8}}=5\left( 15 \right)+4$

$\Rightarrow {{a}_{8}}=79$

So, its ${{8}^{th}}$ term is 79.

14. The fourth term of an AP is 11 and the eighth  term exceeds twice the fourth  term by 5. Find the AP and find the sum of the first 50  terms.

Ans: The fourth term of an AP is 11.

$\Rightarrow {{a}_{4}}=11$

$\Rightarrow a+3d=11$                  ……….(1)

The eighth  term exceeds twice the fourth  term by 5.

$\Rightarrow {{a}_{8}}=2{{a}_{4}}+5$

$\Rightarrow a+7d=2\left( a+3d \right)+5$

$\Rightarrow a+7d=2\left( 11 \right)+5$           [From eq(1)]

$\Rightarrow a+7d=27$                 ………..(2)

Subtract eq (1) from eq (2)-

$a+7d-a-3d=27-11$

$\Rightarrow 4d=16$

$\Rightarrow d=4$

From eq(1):

$a+3\left( 4 \right)=11$

$\Rightarrow a= -1$

Second term:

${{a}_{2}}=a+d$

${{a}_{2}}= -1+4=3$

Third term:

${{a}_{3}}=a+2d$

${{a}_{3}}= -1+8=7$

So, AP is -1, 3, 7, 11……....

The sum of first 50 terms can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2} \left[  2a+\left( n-1 \right)d  \right]$

$\Rightarrow {{S}_{50}}=\dfrac{50}{2} \left[  2\left( -1 \right)+\left( 50-1 \right)4  \right]$

$\Rightarrow {{S}_{50}}=25 \left[  -2+196  \right]$

$\Rightarrow {{S}_{50}}=25 \left[  194  \right]$

$\Rightarrow {{S}_{50}}=4850 $

So, The sum of the first 50 terms is 4850.

Exercise 10(D)

1. Find three numbers in A.P. whose sum is 24 and whose product is 440.

Ans: Consider the first term of the AP is a - d.

The second term of the AP is a.

Third term of the AP is a + d.

The sum of the three terms is 24.

a – d + a + a + d = 24

$\Rightarrow $3a = 24

$\Rightarrow a=8$

The product of the three terms is 440.

$\left( a-d \right)\left( a \right)\left( a+d \right)=440$

$\Rightarrow \left( 8-d \right)\left( 8 \right)\left( 8+d \right)=440$

$\Rightarrow {{8}^{2}}-{{d}^{2}}=55$

$\Rightarrow {{d}^{2}}=64-55=9$

$\Rightarrow d= \pm 3$

If d = 3,

The first term of the AP is a - d = 8 - 3 = 5.

The second term of the AP is a = 8.

Third term of the AP is a + d = 8 + 3 = 11.

If d = -3,

The first term of the AP is a - d = 8 +3 = 11.

The second term of the AP is a = 8.

Third term of the AP is a + d = 8 - 3 = 5.

So, from both the APs we get the same numbers as 5, 8 and 11.

2. Sum of three consecutive terms of an AP is 21 and the sum of their squares is 165. Find these terms.

Ans: Consider the first term of the AP is a - d.

The second term of the AP is a.

Third term of the AP is a + d.

The sum of the three terms is 21.

$a-d+a+a+d=21$

$\Rightarrow 3a=21$

$\Rightarrow a=7$

The sum of their squares is 165.

${{\left( a-d \right)}^{2}}+{{a}^{2}}+{{\left( a+d \right)}^{2}}=165$

$\Rightarrow {{a}^{2}}+{{d}^{2}}-2ad+{{a}^{2}}+{{a}^{2}}+{{d}^{2}}+2ad=165$

$\Rightarrow 3{{a}^{2}}+2{{d}^{2}}=165$

$\Rightarrow 3{{(7)}^{2}}+2{{d}^{2}}=165$

$\Rightarrow 147+2{{d}^{2}}=165$

$\Rightarrow {{d}^{2}}=\dfrac{18}{2}$

$\Rightarrow {{d}^{2}}=$9

$\Rightarrow d= \pm 3$

If d = 3,

The first term of the AP is a - d = 7 - 3 = 4.

The second term of the AP is a = 7.

Third term of the AP is a + d = 7 + 3 = 10.

If d = -3,

The first term of the AP is a - d = 7 + 3 = 10.

The second term of the AP is a = 7.

Third term of the AP is a + d = 7 - 3 = 4.

So, from both the APs we get the same numbers as 4, 7 and 10.

3. The angles of a quadrilateral are in AP with common difference${{20}^{\circ }}$. Find its angles.

Ans: Let the first angle is a and the common difference is ${{20}^{\circ }}$. Therefore, The second angle is $a+{{20}^{\circ }}$.

The third angle is $a+{{40}^{\circ }}$.

The fourth angle is $a+{{60}^{\circ }}$.

We know that sum of all the angles of a quadrilateral is ${{360}^{\circ }}.$

$\Rightarrow a+a+20+a+40+a+60=360$

$\Rightarrow 4a+120=360$

$\Rightarrow 4a=240$

$\Rightarrow a={{60}^{\circ }}$

Therefore, The second angle is $60+20={{80}^{\circ }}$.

The third angle is $60+40={{100}^{\circ }}$.

The fourth angle is $60+60={{120}^{\circ }}$.

4. Divide 96 into four parts which are in AP and the ratio between products of their means to products of their extremes is 15:7.

Ans: Let the four numbers are ‘a-3d’, ‘a-d’, ‘a+d’, ‘a+3d’.

Given that sum of numbers is 96.

$\Rightarrow a-3d+a-d+a+d+a+3d=96$

$\Rightarrow 4a=96$

$\Rightarrow a=24$

The ratio between the products of their means to the product of their extremes is 15:7.

$\Rightarrow \dfrac{\left( a-d \right)\left( a+d \right)}{\left( a-3d \right)\left( a+3d \right)}=\dfrac{15}{7}$

$\Rightarrow \dfrac{{{a}^{2}}-{{d}^{2}}}{{{a}^{2}}-9{{d}^{2}}}=\dfrac{15}{7}$

$\Rightarrow 7\left( {{a}^{2}}-{{d}^{2}} \right)=15\left( {{a}^{2}}-9{{d}^{2}} \right)$

$\Rightarrow 7{{a}^{2}}-7{{d}^{2}}=15{{a}^{2}}-135{{d}^{2}}$

$\Rightarrow 8{{a}^{2}}=128{{d}^{2}}$

$\Rightarrow 8\times {{24}^{2}}=128{{d}^{2}}$

$\Rightarrow {{d}^{2}}=36$

$\Rightarrow d= \pm 6$

If d = 6,

The four numbers are ‘24-3(6)’, ‘24-6’, ‘24+6’, ‘24+3(6)’.

So, the four numbers are ‘6’, ‘18’, ‘30’, ‘42’.

If d = -6,

The four numbers are ‘24-3(-6)’, ‘24+6’, ‘24-6’, ‘24+3(-6)’.

So, the four numbers are ‘42’, ‘30’, ‘18’, ‘6’.

So from both the APs we get the same numbers as ‘6’, ‘18’, ‘30’ and ‘42’.

5. Five numbers are in AP whose sum is 12$\dfrac{1}{2}$and the ratio of the first to the last  term is 2:3.

Ans: Let the five numbers are ‘a-2d’, ‘a-d’, ‘a’ ‘a+d’ and ‘a+2d’.

Given that the sum of numbers is 12$\dfrac{1}{2}$.

$\Rightarrow a-2d+a-d+a+a+d+a+2d=\dfrac{25}{2}$

$\Rightarrow 5a=\dfrac{25}{2}$

$\Rightarrow a=\dfrac{5}{2}$

The ratio of the first to the last  term is 2:3.

$\Rightarrow \dfrac{a-2d}{a+2d}=\dfrac{2}{3}$

$\Rightarrow 3a-6d=2a+4d$

$\Rightarrow a=10d$

$\Rightarrow d=\dfrac{5}{2\times 10}$

$\Rightarrow d=\dfrac{1}{4}$

So the first  term is = $a-2d$

= $\dfrac{5}{2}-2\times \dfrac{1}{4}$=2

So the second  term is = $a-d$

= $\dfrac{5}{2}-\dfrac{1}{4}$= $\dfrac{9}{4}$

So the third  term is = $a=\dfrac{5}{2}$

So the fourth  term is = $a+d$

= $\dfrac{5}{2}+\dfrac{1}{4}$= $\dfrac{11}{4}$

So the fifth  term is = $a+2d$

= $\dfrac{5}{2}+2\times \dfrac{1}{4}$= $3$

6. Split 207 into three parts such that these parts are in AP and the product of two smaller parts is 4623.

Ans:  Let the three numbers are ‘a-d’, ‘a’ and ‘a+d’.

Given that sum of numbers is 207.

$\Rightarrow a-d+a+a+d=207$

$\Rightarrow 3a=207$

$\Rightarrow a=69$

The product of two smaller parts is 4623.

$\left( a-d \right)a=4623$

$\Rightarrow \left( 69-d \right)69=4623$

$\Rightarrow \left( 69-d \right)=67$

$\Rightarrow d=2$

The three numbers are ‘69-2’, ‘69’ and ‘69+2’.

So, the numbers are 67, 69 and 71.

7. The sum of three numbers in AP is 15 and the sum of the squares of the extreme terms is 58. Find the numbers.

Ans: Let the three numbers are ‘a-d’, ‘a’ and ‘a+d’.

Given that sum of numbers is 15.

$\Rightarrow a-d+a+a+d=15$

$\Rightarrow 3a=15$

$\Rightarrow a=5$

The sum of the squares of the extreme terms is 58.

${{(a-d)}^{2}}+{{(a+d)}^{2}}=58$

$\Rightarrow {{a}^{2}}+{{d}^{2}}-2ad+{{a}^{2}}+{{d}^{2}}+2ad=58$

$\Rightarrow 2\left( {{a}^{2}}+{{d}^{2}} \right)=58$

$\Rightarrow {{5}^{2}}+{{d}^{2}}=29$

$\Rightarrow {{d}^{2}}=4$

$\Rightarrow d= \pm 2$

If d = 2

The three numbers are ‘5-2’, ‘5’  and ‘5+2’.

So, the numbers are 3, 5 and 7.

If d = -2

The three numbers are ‘5+2’, ‘5’  and ‘5-2’.

So, the numbers are 3, 5 and 7.

8. Find the four numbers in AP whose sum is 20 and the sum of whose squares is 120.

Ans: Let the four numbers are ‘a-3d’, ‘a-d’, ‘a+d’, ‘a+3d’.

Given that sum of numbers is 20.

$\Rightarrow a-3d+a-d+a+d+a+3d=20$

$\Rightarrow 4a=20$

$\Rightarrow a=5$

The sum of whose squares is 120.

$\Rightarrow {{\left( a-3d \right)}^{2}}+{{\left( a-d \right)}^{2}}+{{\left( a+d \right)}^{2}}+{{\left( a+3d \right)}^{2}}=120$

$\Rightarrow {{a}^{2}}+9{{d}^{2}}-6ad+{{a}^{2}}+{{d}^{2}}-2ad+{{a}^{2}}+{{d}^{2}}+2ad+{{a}^{2}}+9{{d}^{2}}+6ad=120$

$\Rightarrow 4{{a}^{2}}+20{{d}^{2}}=120$

$\Rightarrow {{5}^{2}}+5{{d}^{2}}=30$

$\Rightarrow 5{{d}^{2}}=5$

$\Rightarrow d= \pm 1$

If d = 1,

So, the four numbers are ‘5-3’, ‘5-1’, ‘5+1’, ‘5+3’.

And if d = -1,

So, the four numbers are ‘5+3’, ‘5+1’, ‘5-1’, ‘5-3’.

So, the numbers are 2, 4, 6 and 8 from both the APs.

9. Insert one arithmetic mean between 3 and 13.

Ans: The two numbers are 3 and 13.

Consider ‘a’ to be the arithmetic mean between 3 and 13.

The arithmetic mean can be calculated as follows,

$\Rightarrow a=\dfrac{3+13}{2}$

$\Rightarrow a=\dfrac{16}{2}$

$\Rightarrow a=8$

The arithmetic mean between 3 and 13 is 8.

10. The angles of a polygon are in AP with common difference ${{5}^{\circ }}.$If the smallest angle is ${{120}^{\circ }},$find the number of sides of the polygon.

Ans: Consider ‘n’ to be the number of sides in a polygon.

The common difference is 5.

The smaller angle is 120.

The number of sides in a polygon can be calculated as follows,

We know that if there n sides of a polygon then sum of all the angles is given by$\left( n-2 \right)\times 180.$

From the given we can form AP as:

120, 125, 130, …....

Sum of n terms of an AP is given by the following formula:

$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[  2a +\left( n-1 \right)d  \right]$

$\Rightarrow \left( n-2 \right)\times 180=\dfrac{n}{2}\left[  2\left( 120 \right) +\left( n-1 \right)5  \right]$

$\Rightarrow 360n-720=n\left[  240 +5n-5  \right]$

$\Rightarrow 360n-720=235n +5{{n}^{2}}$

$\Rightarrow 0=720+235n-360n +5{{n}^{2}}$

$\Rightarrow 5{{n}^{2}}-125n+720=0$

$\Rightarrow {{n}^{2}}-25n+144=0$

$\Rightarrow {{n}^{2}}-16n-9n+144=0$

$\Rightarrow n\left( n-16 \right)-9\left( n-16 \right)=0$

$\Rightarrow \left( n-16 \right)\left( n-9 \right)=0$

$\Rightarrow n=9 \And  16$

The number of sides in a polygon can be 9 or 16.

11. $\dfrac{1}{a}, \dfrac{1}{b}$ and $\dfrac{1}{c}$ are in AP. Show that bc, ac and ab are also in AP.

Ans: Given $\dfrac{1}{a}, \dfrac{1}{b}$and $\dfrac{1}{c}$are in AP.

The arithmetic mean can be written as:

$\Rightarrow 2\times \dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}$                ………(1)

We need to show that bc, ac and ab are also in AP.

Divide each term by abc. Then,

$\dfrac{bc}{abc}, \dfrac{ac}{abc} and \dfrac{ab}{abc}$

$\Rightarrow \dfrac{1}{a}, \dfrac{1}{b}and \dfrac{1}{c}$                      ……..(2)

From eq(1) we know that if the following condition satisfies then these  terms are in AP.

$2\times \dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}$

So eq(2) satisfies the condition that means

$\dfrac{bc}{abc}, \dfrac{ac}{abc} and \dfrac{ab}{abc}$ are in AP.

So, we can write it as: bc, ac and ab are also in AP.

Hence proved.

Exercise 10(E)

1. Two cars start together in the same direction from the same place. The first car goes at a uniform speed of 10 km/h. The second car goes at a speed of 8 km/h in the first hour and thereafter increasing the speed by 0.5 km/h each succeeding hour. After how many hours will the two cars meet?

Ans: The uniform speed of the first car is 10 km/h.

The speed of the second car in the first hour is 8 km/h.

The speed of the car is increasing by 0.5 km/h.

Consider the two cars meeting after n hours.

So, the distance covered by first car = $10\times n km$

We know that distance = speed $\times  $time

Time = 1 hour then distance = speed

Distance covered by second car forms an AP as follows,

8, 8.5, 9, 9.5………

So, Sum of n terms of an AP is given by the following formula:

$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[  2a +\left( n-1 \right)d  \right]$

$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[  2\left( 8 \right) +\left( n-1 \right)\left( 0.5 \right)  \right]$

As both cars meet after n hours so both will cover equal distance after n hours.

$\Rightarrow 10\times n=\dfrac{n}{2}\left[  2\left( 8 \right) +\left( n-1 \right)\left( 0.5 \right)  \right]$

$\Rightarrow 10\times 2=\left[  16 +\left( \dfrac{n}{2}-\dfrac{1}{2} \right)  \right]$

$\Rightarrow \dfrac{n}{2}=\dfrac{9}{2}$

$\Rightarrow n=9$

So, After 9 hours the two cars will meet.

2. A sum of Rs. 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize Rs 20 less than its preceding; find the value of each of the prizes.

Ans: The number of prizes distributed is n = 7.

The common difference is 20.

The sum of prizes is 700.

$\Rightarrow {{S}_{7}}=\dfrac{7}{2}\left[ 2a+6\times 20 \right]$

$\Rightarrow 700=\dfrac{7}{2}\left[ 2a+6\times 20 \right]$

$\Rightarrow \dfrac{700\times 2}{7}=\left[ 2a+120 \right]$

$\Rightarrow 200-120 = 2a$

$\Rightarrow a=\dfrac{80}{2}=40$

The prize given to the student who got seventh position is 40.

The prize given to the student who got sixth position is 60.

The prize given to the student who got fifth position is 80.

The prize given to the student who got fourth position is 100.

The prize given to the student who got third position is 120.

The prize given to the student who got second position is 140.

The prize given to the student who got first position is 160.

3. An article can be bought by paying Rs 28000 at once or by making 12 monthly installments. If the first instalment is Rs 3000 and every other instalment is 100 less than the previous one, find:

i) amount of installment paid in the ${{9}^{th}}$month.

ii) total amount paid in the instalment scheme.

Ans: i) The first payment is 3000.

The common difference is -100.

The ninth payment can be calculated as follows,

$\Rightarrow {{a}_{9}}=a+\left( 9-1 \right)d$

$\Rightarrow {{a}_{9}}=3000+\left( 8 \right)\left( -100 \right)$

$\Rightarrow {{a}_{9}}=3000-800$

$\Rightarrow {{a}_{9}}=2200$

ii) The total amount can be calculated as follows,

$\Rightarrow {{S}_{12}}=\dfrac{12}{2}\left[ 2\times 3000+\left( 12-1 \right)\times \left( -100 \right) \right]$

$\Rightarrow {{S}_{12}}=6\left[ 6000-1100 \right]$

$\Rightarrow {{S}_{12}}=6\left[ 4900 \right]$

$\Rightarrow {{S}_{12}}=29400$

4. A manufacturer of TV sets produces 600 units in the third year and 700 units in the ${{7}^{th}}$ year. Assuming that the production increases uniformly by a fixed number year, find:

i) the production in the first year.

ii) the production in ${{10}^{th}}$year.

iii) the total production in 7 years.

Ans: 

i) A manufacturer produces 600 units in the third year.

$\Rightarrow {{a}_{3}}=a+2d =600$

A manufacturer produces 700 units in the seventh year.

$\Rightarrow {{a}_{7}}=a+6d =700$

Subtract third term from seventh term.

$\Rightarrow a+6d-a-2d=700-600$

$\Rightarrow 4d=100$

$\Rightarrow d=25$

The first year manufactured units can be obtained as follows using third  term,

$\Rightarrow a+2\left( 25 \right)=600$

$\Rightarrow a=600-50=550$

ii) the production in ${{10}^{th}}$year can be calculated as follows,

$\Rightarrow {{a}_{10}}=a+9d $

$\Rightarrow {{a}_{10}}=550+8\left( 25 \right) $

$\Rightarrow {{a}_{10}}=550+200$

$\Rightarrow {{a}_{10}}=750$

iii) the total production in 7 years can be calculated as follows,

$\Rightarrow {{S}_{7}}=\dfrac{7}{2}\left[ 2\times 550+\left( 7-1 \right)\times \left( 25 \right) \right]$

$\Rightarrow {{S}_{7}}=\dfrac{7}{2}\left[ 1100+150 \right]$

$\Rightarrow {{S}_{7}}=\dfrac{7}{2}\left[ 1250 \right]$

$\Rightarrow {{S}_{7}}=7\times 625$

$\Rightarrow {{S}_{7}}=4375$

5. Mr. Gupta repays her total loan of Rs. 1,18,000 by paying installments every month. If the instalment for the first month is Rs 1000 and it increases by Rs 100 every month, what amount will she pay as the ${{30}^{th}}$instalment of loan? What amount of loan she still has to pay after ${{30}^{th}}$instalment?

Ans: The first month installment is 1000.

The common difference is 100.

The ${{30}^{th}}$instalment can be calculated as follows,

${{a}_{30}}=a+\left( 30-1 \right)d$

${{a}_{30}}=1000+\left( 29 \right)100$

${{a}_{30}}=3900$

The total amount paid till ${{30}^{th}}$ instalment can be calculated as follows,

$\Rightarrow {{S}_{30}}=\dfrac{30}{2}\left[ 2\times 1000+\left( 30-1 \right)\times \left( 100 \right) \right]$

$\Rightarrow {{S}_{30}}=15\left[ 2000+2900 \right]$

$\Rightarrow {{S}_{30}}=15\left[ 4900 \right]$

$\Rightarrow {{S}_{30}}=73500$

The total amount that has to be paid can be calculated as follows,

Amount =118000 - 73500

= 44500

Exercise 10(F)

1. The ${{6}^{th}}$ term of an AP is 16 and the ${{14}^{th}}$ term is 32. Determine the ${{36}^{th}}$ term.

Ans: The ${{6}^{th}}$ term of an AP is 16.

$\Rightarrow {{a}_{6}}=a+5d$

$\Rightarrow 16=a+5d$                     ………(1)

The ${{14}^{th}}$ term of an AP is 32.

$\Rightarrow {{a}_{14}}=a+13d$

$\Rightarrow 32=a+13d$                    ……….(2)

Subtract eq(1) from eq(2)-

$\Rightarrow a+13d-a-5d=32-16$

$\Rightarrow 8d=16$

$\Rightarrow d=2$

So from eq(1)-

$16=a+5\left( 2 \right)$

$\Rightarrow a=16-10=6$

We will calculate ${{36}^{th}}$ term as follows, 

$\Rightarrow {{a}_{36}}=a+35d$

$\Rightarrow {{a}_{36}}=6+35\left( 2 \right)$

$\Rightarrow {{a}_{36}}=76$

2. If the third and ${{9}^{th}}$  terms of an AP be 4 and -8 respectively. Find which  term is zero?

Ans: The ${{3}^{th}}$ term of an AP is 4.

$\Rightarrow {{a}_{3}}=a+2d$

$\Rightarrow 4=a+2d$                   …………(1)

The ${{9}^{th}}$ term of an AP is -8.

$\Rightarrow {{a}_{9}}=a+8d$

$\Rightarrow -8=a+8d$                …………..(2)

Subtract eq(1) from eq(2)-

$\Rightarrow a+8d-a-2d=-8-4$

$\Rightarrow 6d=-12$

$\Rightarrow d=-2$

So from eq(1)-

$4=a+2\left( -2 \right)$

$\Rightarrow a=4+4=8$

Let ${{n}^{th}}$ term be zero.

$\Rightarrow {{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 0=8+\left( n-1 \right)\left( -2 \right)$

$\Rightarrow -8=\left( n-1 \right)\left( -2 \right)$

$\Rightarrow 4=\left( n-1 \right)$

$\Rightarrow n=5$

3. An AP consists of 50 terms of which ${{3}^{rd}}$ term is 12 and the last  term is 106. Find the ${{29}^{th}}$ term of the AP.

Ans: The ${{3}^{th}}$ term of an AP is 12.

$\Rightarrow {{a}_{3}}=a+2d$

$\Rightarrow 12=a+2d$               ……..(1)

The last/${{50}^{th}}$ term of an AP is 106.

$\Rightarrow {{a}_{50}}=a+49d$

$\Rightarrow 106=a+49d$            ………..(2)

Subtract eq(1) from eq(2)-

$\Rightarrow a+49d-a-2d=106-12$

$\Rightarrow 47d=94$

$\Rightarrow d=2$

So from eq(1)-

$12=a+2\left( 2 \right)$

$\Rightarrow a=12-4=8$

The ${{29}^{th}}$ term of the AP can be calculated as follows,

$\Rightarrow {{a}_{29}}=a+\left( 29-1 \right)d$

$\Rightarrow {{a}_{29}}=8+28\left( 2 \right)$

$\Rightarrow {{a}_{29}}=64$

4. Find the arithmetic mean of:

i) -5 and 41

ii) 3x-2y and 3x+2y

iii) ${{\left( m+n \right)}^{2}} and {{\left( m-n \right)}^{2}}$

Ans: 

i) The numbers are -5 and 41.

The arithmetic mean can be calculated as follows,

$\Rightarrow AM=\dfrac{-5+41}{2}$

$\Rightarrow AM=\dfrac{36}{2}$

$\Rightarrow AM=18$

ii) The numbers are 3x-2y and 3x+2y.

The arithmetic mean can be calculated as follows,

$\Rightarrow AM=\dfrac{3x-2y+3x+2y}{2}$

$\Rightarrow AM=\dfrac{6x}{2}$

$\Rightarrow AM=3x$

iii) The numbers are ${{\left( m+n \right)}^{2}} and {{\left( m-n \right)}^{2}}$.

The arithmetic mean can be calculated as follows,

$\Rightarrow AM=\dfrac{{{\left( m+n \right)}^{2}}+{{\left( m-n \right)}^{2}}}{2}$

$\Rightarrow AM=\dfrac{{{m}^{2}}+{{n}^{2}}+2mn+{{m}^{2}}+{{n}^{2}}-2mn}{2}$

$\Rightarrow AM={{m}^{2}}+{{n}^{2}}$

5. Find the sum of the first 10 terms of the AP: 4, 6, 8 …....

Ans: The given AP is 4, 6, 8 ….

The first term is 4.

The common difference can be calculated as follows,

$\Rightarrow d={{a}_{2}}-{{a}_{1}}=6-4=2$

So, the sum of the first 10 terms of the AP:

$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

$\Rightarrow {{S}_{10}}=\dfrac{10}{2}\left[ 2\left( 4 \right)+\left( 10-1 \right)2 \right]$

$\Rightarrow {{S}_{10}}=5\times \left[ 8+18 \right]$

$\Rightarrow {{S}_{10}}=5\times 26$

$\Rightarrow {{S}_{10}}=130$

6. Find the sum of the first 20 terms of an AP whose first  term is 3 and the last  term is 57.

Ans: The first term of the AP is 3.

The last term of the AP is 57.

The sum of first 20 terms can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2} \left[  a+l  \right]$

$\Rightarrow {{S}_{20}}=\dfrac{20}{2} \left[  3+57  \right]$

$\Rightarrow {{S}_{20}}=10 \left[  60  \right]$

$\Rightarrow {{S}_{20}}=600$

7. How many terms of the series 18 + 15 + 12 ….... when added together will give 45?

Ans: The given series is 18 + 15 + 12 …....

Here the first  term is 18.

The common difference is 15 - 18 = -3

Let the sum of n  terms of the AP is 45.

The sum of first n  terms can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2} \left[  2a+\left( n-1 \right)d  \right]$

$\Rightarrow 45=\dfrac{n}{2} \left[  2\left( 18 \right)+\left( n-1 \right)\left( -3 \right) \right]$

$\Rightarrow 90= \left[  36n-3n\left( n-1 \right) \right]$

$\Rightarrow 90= \left[  36n-3{{n}^{2}}+3n \right]$

$\Rightarrow 3{{n}^{2}}-39n+90=0$

$\Rightarrow {{n}^{2}}-13n+30=0$

$\Rightarrow {{n}^{2}}-10n-3n+30=0$

$\Rightarrow n\left( n-10 \right)-3\left( n-10 \right)=0$

$\Rightarrow \left( n-3 \right)\left( n-10 \right)=0$

So $\Rightarrow n=3 \And  10$

So we can say that the sum of 3 terms or  10  terms will give us 45.

8. The ${{n}^{th}}$  term of a sequence is 8 - 5n. Show that the sequence is an AP.

Ans: The ${{n}^{th}}$  term of a sequence is 8 - 5n.

Substitute n = 1,

${{a}_{1}}=8-5\left( 1 \right)=3$

Substitute n = 2,

${{a}_{2}}=8-5\left( 2 \right)= -2$

Substitute n = 3,

${{a}_{3}}=8-5\left( 3 \right)= -7$

Now, we know that a sequence is in AP if the common difference between consecutive terms is equal.

The common difference = ${{a}_{2}}-{{a}_{1}}= -2 - \left( 3 \right) = -5$

The common difference = ${{a}_{3}}-{{a}_{2}}= -7-\left( -2 \right)= -5$

So, the common difference is equal so we can say that the given sequence is an AP.

9. Find the general term(${{n}^{th}} term)$and ${{23}^{rd}}$ terms of the sequence 3, 1, -1, -3 ....

Ans: The given sequence is 3, 1, -1, -3 ....

The common difference = ${{a}_{2}}-{{a}_{1}}= 1 - \left( 3 \right) = -2$

The common difference = ${{a}_{3}}-{{a}_{2}}= -1-\left( 1 \right)= -2$

So the common difference is equal so we can say that the given sequence is an AP with first term 3 and common difference (-2).

So the general term(${{n}^{th}} term)$can be calculated as follows,

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow {{a}_{n}}=3+\left( n-1 \right)\left( -2 \right)$

$\Rightarrow {{a}_{n}}=3-2n+2$

$\Rightarrow {{a}_{n}}=5-2n$

So now we can find ${{23}^{rd}}$ term as follows,

$\Rightarrow {{a}_{23}}=5-2\left( 23 \right)$

$\Rightarrow {{a}_{23}}= 5-46$

$\Rightarrow {{a}_{23}}= -41$

10. Which term of the sequence 3, 8, 13 …… is 78?

Ans: The given sequence is 3, 8, 13 ....

The common difference = ${{a}_{2}}-{{a}_{1}}= 8 - \left( 3 \right) = 5$

The common difference = ${{a}_{3}}-{{a}_{2}}= 13-\left( 8 \right)= 5$

So the common difference is equal so we can say that the given sequence is an AP with first term 3 and common difference 5.

Let the ${{n}^{th}} \text{term} $of the sequence be 78.

So the ${{n}^{th}} term$can be calculated as follows,

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 78=3+\left( n-1 \right)\left( 5 \right)$

$\Rightarrow 78-3=5n-5$

$\Rightarrow 75+5=5n$

$\Rightarrow 80=5n$

$\Rightarrow n=\dfrac{80}{5}=16$

So, the ${{16}^{th}}$ term of the sequence is 78.

11. Is -150 a term of 11, 8, 5, 2 ……....?

Ans: The given sequence is 11, 8, 5 ....

The common difference = ${{a}_{2}}-{{a}_{1}}= 8 - \left( 11 \right) = -3$

The common difference = ${{a}_{3}}-{{a}_{2}}= 5-\left( 8 \right)= -3$

So the common difference is equal so we can say that the given sequence is an AP with first term 11 and common difference -3.

Let the ${{\text{n}}^{\text{th}}}\text{ }\!\! \!\!\text{ term}$ of the sequence is -150.

So, the ${{n}^{th}} term$can be calculated as follows,

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow -150=11+\left( n-1 \right)\left( -3 \right)$

$\Rightarrow -161=-3n+3$

$\Rightarrow -164=-3n$

$\Rightarrow n=\dfrac{-164}{-3}$

So here we get the value of n as a fraction. So -150 is not a term of 11, 8, 5, 2 ……....

12. How many two-digit numbers are divisible by 3?

Ans: The first two digit number which is divisible by 3 is 12.

The second two digit number which is divisible by 3 is 15.

The third two digit number which is divisible by 3 is 18.

The last two digit number which is divisible by 3 is 99.

So the sequence is 12, 15, 18 ……....99

Here the first  term is 12 and the common difference is 15 - 12 = 3.

Let the ${{n}^{th}}$ term of the sequence be 99.

So the ${{n}^{th}} term$can be calculated as follows,

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 99=12+\left( n-1 \right)\left( 3 \right)$

$\Rightarrow 87=3n-3$

$\Rightarrow 90=3n$

$\Rightarrow n=\dfrac{90}{3}$

$\Rightarrow n=30$

So, there are 30 two digit numbers which are divisible by 3.

13. How many multiples of 4 lie between 10 and 250?

Ans: The first number between 10 and 250 which is multiples of 4 is 12.

The second number between 10 and 250 which is multiples of 4 is 16.

The third number between 10 and 250 which is multiples of 4 is 20.

The last number between 10 and 250 which is multiples of 4 is 248.

So the sequence is 12, 16, 20 ……....248

Here the first  term is 12 and the common difference is 16 - 12 = 4.

Let the ${{n}^{th}}$ term of the sequence is 248.

So the ${{n}^{th}} term$can be calculated as follows,

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 248=12+\left( n-1 \right)\left( 4 \right)$

$\Rightarrow 236=4n-4$

$\Rightarrow 240=4n$

$\Rightarrow n=\dfrac{240}{4}$

$\Rightarrow n=60$

So, there are 60 multiples of 4 lying between 10 and 250.

14. The sum of the ${{4}^{th}}$ term and the ${{8}^{th}}$ term of an AP is 24 and the sum of the ${{6}^{th}}$ term and the ${{10}^{th}}$ term is 44. Find the first three terms of an AP.

Ans: The sum of the ${{4}^{th}}$ term and the ${{8}^{th}}$ term of an AP is 24.

$\Rightarrow {{a}_{4}}+{{a}_{8}}=24$

We know That:${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow a+3d+a+7d=24$

$\Rightarrow 2a+10d=24$

$\Rightarrow a+5d=12$                  ……..(1)

The sum of the ${{6}^{th}}$ term and the ${{10}^{th}}$ term is 44.

$\Rightarrow {{a}_{6}}+{{a}_{10}}=44$

$\Rightarrow a+5d+a+9d=44$

$\Rightarrow 2a+14d=44$

$\Rightarrow a+7d=22$                     …….(2)

Subtract eq(1) from eq(2)-

$\Rightarrow a+7d-a-5d=22-12$

$\Rightarrow 2d=10$

$\Rightarrow d=5$

From eq(1) :

$\Rightarrow a+5\left( 5 \right)=12$

$\Rightarrow a=12-25= -13$

Second Term:

$\Rightarrow {{a}_{2}}=a+d$

$\Rightarrow {{a}_{2}}= -13+5$

$\Rightarrow {{a}_{2}}= -8$

Third Term:

$\Rightarrow {{a}_{3}}=a+2d$

$\Rightarrow {{a}_{3}}= -13+2\times 5$

$\Rightarrow {{a}_{3}}= -3$

15. The sum of the first 14 terms of an AP is 1050 and its ${{14}^{th}}$ term is 140. Find its ${{20}^{th}}$  term.

Ans: The ${{14}^{th}}$ term is 140.

$\Rightarrow {{a}_{14}}=a+13d$

$\Rightarrow 140=a+13d$                        ……….(1)

 The sum of the first 14 terms of an AP is 1050.

${{S}_{14}}=\dfrac{14}{2} \left[  a+l  \right]$

$\Rightarrow 1050=7 \left[  a+140  \right]$

$\Rightarrow \dfrac{1050}{7}=a+140$

$\Rightarrow 150-140=a$

$\Rightarrow a=10$

From eq(1):

$\Rightarrow 140=10+13d$

$\Rightarrow d=\dfrac{130}{13}=10$

The ${{20}^{th}}$ term can be calculated as follows,

$\Rightarrow {{a}_{20}}=a+19d$

$\Rightarrow {{a}_{20}}=10+19\left( 10 \right)$

$\Rightarrow {{a}_{20}}=200$

16. The ${{25}^{th}}$ term of an AP exceeds its ${{9}^{th}}$ term by 16. Find its common difference.

Ans: The ${{25}^{th}}$ term of an AP exceeds its ${{9}^{th}}$ term by 16.

Let the first term of an AP is a and the common difference is d.

$\Rightarrow {{a}_{25}}={{a}_{9}}+16$

$\Rightarrow a+24d=a+8d+16$

$\Rightarrow 16d=16$

$\Rightarrow d=1$

17. For an AP, Show that:

${{\left( m+n \right)}^{th}}$ term + ${{\left( m-n \right)}^{th}}$ term = $2\times {{m}^{th}}$ term

Ans: Let the first term of an AP is a and the common difference is d.

${{\left( m+n \right)}^{th}}$ term can be calculated as follows,

${{a}_{m+n}}=a+\left( m+n-1 \right)d$

${{\left( m-n \right)}^{th}}$ term can be calculated as follows,

${{a}_{m-n}}=a+\left( m-n-1 \right)d$

${{\left( m \right)}^{th}}$ term can be calculated as follows,

${{a}_{m}}=a+\left( m-1 \right)d$

Now taking LHS,

${{\left( m+n \right)}^{th}}$ term + ${{\left( m-n \right)}^{th}}$ term

$={{a}_{m+n}}$+${{a}_{m-n}}$

=$a+\left( m+n-1 \right)d$+$a+\left( m-n-1 \right)d$

$=a+md+nd-d+a+md-nd-d$

$=2a+2md-2d$

$=2\left( a+md-d \right)$

$=2\left[ a+\left( m-1 \right)d \right]$

$=2\times {{a}_{m}}=RHS$

Hence LHS = RHS

Hence proved.

18. If the ${{n}^{th}}$ term of an AP 58, 60, 62 ….... is equal to the ${{n}^{th}}$ term of an AP -2, 5, 12 ….... find the value of n.

Ans: The first AP is 58, 60, 62 ……

Here the first term is 58 and the common difference is (60 - 58 = 2).

So, the ${{n}^{th}}$ term of an AP can be calculated as follows,

$\Rightarrow {{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow {{a}_{n}}=58+\left( n-1 \right)2$

$\Rightarrow {{a}_{n}}=58+2n-2$

$\Rightarrow {{a}_{n}}=56+2n$

 The second AP is -2, 5, 12 ……

Here the first term is -2 and the common difference is [5 - (-2) = 7].

So, the ${{n}^{th}}$ term of an AP can be calculated as follows,

$\Rightarrow {{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow {{a}_{n}}= -2+\left( n-1 \right)7$

$\Rightarrow {{a}_{n}}= -2+7n-7$

$\Rightarrow {{a}_{n}}=7n-9$

The ${{n}^{th}}$ term of an AP 58, 60, 62 ….... is equal to the ${{n}^{th}}$ term of an AP -2, 5, 12 ….... 

$\Rightarrow 56+2n=7n-9$

$\Rightarrow 56+9=5n$

$\Rightarrow n=\dfrac{65}{5}$

$\Rightarrow n=13$

19. Which term of the AP 105, 101, 97 …… is the first negative  term?

Ans: The given AP 105, 101, 97 ……

Here the first term is 105 and the second term is 101.

So, the common difference is 101 - 105 = -4.

Let the ${{n}^{th}}$  term of an AP is the first negative term.

$\Rightarrow {{a}_{n}}<0$

$\Rightarrow a+\left( n-1 \right)d<0$

$\Rightarrow 105+\left( n-1 \right)\left( -4 \right)<0$

$\Rightarrow 105-4n+4<0$

$\Rightarrow 4n >109$

\[\Rightarrow n > \dfrac{109}{4}\]

$\Rightarrow n > 27.25$

So from here we can say that the first negative  term is ${{28}^{th}}$ term.

20. How many three digit numbers are divisible by 7?

Ans: The first three digit number which is divisible by 7 is 105.

The second three digit number which is divisible by 7 is 112.

The third three digit number which is divisible by 7 is 119.

The last three digit number which is divisible by 7 is 994.

So the sequence is 105, 112, 119 ……....994

Here the first  term is 105 and the common difference is 112 - 105 = 7.

Let the ${{n}^{th}}$ term of the sequence is 994.

So the ${{n}^{th}} term$can be calculated as follows,

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 994=105+\left( n-1 \right)\left( 7 \right)$

$\Rightarrow 889=7n-7$

$\Rightarrow 896=7n$

$\Rightarrow n=\dfrac{896}{7}$

$\Rightarrow n=128$

So, there are 128 three digit numbers which are divisible by 7.

21. Divide 216 into three parts which are in AP and the product of two smaller parts is 5040.

Ans: Let the three parts are $a-d, a, a+d$ which are in AP.

$\Rightarrow a-d+a+a+d=216$

$\Rightarrow 3a=216$

$\Rightarrow a=72$

The product of two smaller parts is 5040.

$\Rightarrow \left(  a-d  \right)\times a=5040$

$\Rightarrow \left(  72-d  \right)\times 72=5040$

$\Rightarrow \left(  72-d  \right)=\dfrac{5040}{72}$

$\Rightarrow 72-d=70$

$\Rightarrow d=2$

So $a-d, a, a+d$ are (72 - 2, 72, 72 + 2) respectively.

So, three parts of 216 are 70, 72, 74.

22. Can $2{{n}^{2}}-7$be the ${{n}^{th}}$ term of an AP? Explain.

Ans: Let $2{{n}^{2}}-7$is the ${{n}^{th}}$ term of an AP.

Substitute n = 1,

$\Rightarrow {{a}_{1}}=2{{(1)}^{2}}-7= -5$

Substitute n = 2,

$\Rightarrow {{a}_{2}}=2{{(2)}^{2}}-7= 1$

Substitute n = 3,

$\Rightarrow {{a}_{3}}=2{{(3)}^{2}}-7= 11$

As we assumed that $2{{n}^{2}}-7$is the ${{n}^{th}}$ term of an AP then the common difference will be equal.

$\Rightarrow {{a}_{3}}-{{a}_{2}}={{a}_{2}}-{{a}_{1}}$

$\Rightarrow 11-1=1-\left( -5 \right)$

$\Rightarrow 10=6$ which can’t be true. 

So, we can say that our assumption is wrong. So $2{{n}^{2}}-7$can’t be the ${{n}^{th}}$ term of an AP.

23. Find the sum of an AP: 14, 21, 28 …….... 168.

Ans: The given AP is 14, 21, 28 …….... 168

Here the first  term is 14.

The common difference is 21 - 14 = 7

Let there are n  terms in the given AP and ${{n}^{th}}$ term can be calculated as follows

${{a}_{n}}=a+\left( n-1 \right)d$

$\Rightarrow 168=14+\left( n-1 \right)7$

$\Rightarrow 154=\left( n-1 \right)7$

$\Rightarrow n-1=22$

$\Rightarrow n=23$

The sum of the n  terms can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2} \left[  a+l  \right]$

$\Rightarrow {{S}_{23}}=\dfrac{23}{2} \left[  14+168  \right]$

$\Rightarrow {{S}_{23}}=\dfrac{23}{2} \left[  182  \right]$

$\Rightarrow {{S}_{23}}=\left[  23\times 91 \right] $

$\Rightarrow {{S}_{23}}=2093$

24. The first term of an AP is 20 and the sum of its first seven  terms is 2100; find its ${{31}^{st}}$  term.

Ans: The first term of an AP is 20. 

Let the common difference is d.

The sum of the n terms can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2} \left[  2a+\left( n-1 \right)d  \right]$

For n = 7,

$\Rightarrow 2100=\dfrac{7}{2} \left[  2\left( 20 \right)+\left( 7-1 \right)d  \right]$

$\Rightarrow \dfrac{4200}{7}=40+6d$

$\Rightarrow 600-40=6d$

$\Rightarrow d=\dfrac{560}{6}=\dfrac{280}{3}$

So, the ${{31}^{st}}$ term can be calculated as follows,

$\Rightarrow {{a}_{31}}= a+30\left( d \right)$

$\Rightarrow {{a}_{31}}= 20+30\times \dfrac{280}{3}$

$\Rightarrow {{a}_{31}}= 20+2800=2820$

25. Find the sum of last 8 terms of the AP:

-12, -10, -8, ……, 58.

Ans: Here the given AP is -12, -10, -8, ……, 58.

So, after reversing the AP,

AP: 58, 56, 54 …… -8, -10, -12.

So now we need the find sum of the first 8 terms of the reversed AP.

Here the first term is 58 and the second term is 56.

The common difference is 56 - 58 = -2

The sum of the n terms can be calculated as follows,

${{S}_{n}}=\dfrac{n}{2} \left[  2a+\left( n-1 \right)d  \right]$

For n = 8,

$\Rightarrow {{S}_{8}}=\dfrac{8}{2} \left[  2\left( 58 \right)+\left( 8-1 \right)\left( -2 \right) \right]$

$\Rightarrow {{S}_{8}}=4 \left[  116+\left( 7 \right)\left( -2 \right) \right]$

$\Rightarrow {{S}_{8}}=4 \left[  102 \right]$

$\Rightarrow {{S}_{8}}=408...$

Download Class 10 ICSE Solutions for Chapter 10 - Arithmetic Progression Free PDF from Vedantu

Introduction of the Chapter Arithmetic Progression

Arithmetic progression is a progression in which every next term has the same difference as the previous one. We call it in short AP. In other terms, the difference between the previous term and the next term will be common. This difference is called a common difference. 

The common pattern of an  Arithmetic progression is :

a, a+d, a+2d and so on.

Here "a" is the first term of the AP and "d" is the common difference of the AP.

• a = first term of AP

• d = difference between two consecutive terms

You can find consecutive terms of an AP as

• So, The first term of the AP = a

• The second term of AP = a+d

• The third term of the AP = a+ 2d

• The fourth term of the AP = a+3d

• the nth term of the AP = a+ (n-1)d

The above formula is also used to find the general term of AP.

For example, if a has had a first term (a) = 10 and common difference (d) = 2

Then its 8th term would be 

= a+(8-1)d

= a+7d

= 10+ 7*2

= 24

There are two types of AP's

1. Finite AP 

Finite AP is the AP that contains finite words. The number of terms in the finite AP is fixed.

For example, 0,2,4,6,8.

2. Infinite AP 

Infinite AP is the AP that contains infinite words. It is generally represented by continuous dots after the last term.

For example, 0,2,4,6,8…

Sum of an AP

The Sum of n terms of an AP is = n/2( 2a + (n-1)d Or n/2( a+l); where 

• n= no. of terms

• l= last term

• d= common difference

• a= first term

To Find nth Term Using Some Formula

You can also find the nth term of an AP using the formulas of the sum. The nth term of the AP = Sum of n terms - Sum of (n-1) terms

Arithmetic Mean 

If their numbers a, A, and b are in AP, then the arithmetic mean of the AP would be A. In other terms, it is the mid-value for two terms.

The arithmetic mean of a and b is = a + b / 2.

Properties of an AP 

1. Property 1: If you add or subtract the same number from each term of an AP, then the resulting series would also be in AP.

For example, 2, 5, 8, 11 are in AP.

Now adding 2 in all terms resulting in series, 4, 7, 0  13 is also in AP as all of the terms now have a common difference of 3.

Now subtracting 1 in all terms resulting in series, 1, 4, 7, 10  is also in AP as all of the terms now have a common difference of 1.

2. Property 2: If each term of an AP is multiplied or divided by the same no. then, the new series will also be in AP.

For example, 2, 5, 8, 11 are in AP.

Now multiplying 2 in all terms resulting in series, 4, 10, 16, 22 is also in AP as all of the terms now have a common difference of 6.