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) ${\displaystyle \frac{\text{ 1 }}{\text{2}}}\text{ ,}{\displaystyle \frac{\text{ 1 }}{\text{3}}}\text{ ,}{\displaystyle \frac{\text{ 1 }}{\text{4}}}\text{ ,}{\displaystyle \frac{\text{ 1 }}{\text{5}}}......$   

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 ${\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{5}}......$

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

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,

$a_{15}=(2\times 15-3)$

$a_{15}=(30-3)$

$a_{15}=\left(27\right)$

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,

$a_1=(2\times 1+3)$

$a_1=(2+3)$

$a_1=\left(5\right)$

The first  term of the AP is 5.

Substitute n = 2,

$a_2=(2\times 2+3)$

$a_2=(4+3)$

$a_2=\left(7\right)$

The second  term of the AP is 7.

Substitute n = 3,

$a_3=(2\times 3+3)$

$a_3=(6+3)$

$a_3=\left(9\right)$

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.

$a_n=a+(n-1)d$

Substitute 24 for n.

$\Rightarrow a_{24}=a+(24-1)d$

$\Rightarrow a_{24}=12+(24-1)(-2)$

$\Rightarrow a_{24}=12-46$

$\Rightarrow a_{24}=-34$

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

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

${\displaystyle \frac{1}{2}},1,{\displaystyle \frac{3}{2}}..........$

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

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

The difference between consecutive  terms can be calculated as follows.

$a_2-a_1$= 1 - ${\displaystyle \frac{1}{2}}$ = ${\displaystyle \frac{1}{2}}$ , $a_3-a_2={\displaystyle \frac{3}{2}}-1={\displaystyle \frac{1}{2}}$

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

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

$a_n=a+(n-1)d$

Substitute 30 for n.

$\Rightarrow a_{30}=a+(30-1)d$

$\Rightarrow a_{30}={\displaystyle \frac{1}{2}}+(30-1)\left({\displaystyle \frac{1}{2}}\right)$

$\Rightarrow a_{30}={\displaystyle \frac{1}{2}}+{\displaystyle \frac{29}{2}}$

$\Rightarrow a_{30}={\displaystyle \frac{30}{2}}$

$\Rightarrow a_{30}=15$

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.

$a_n=a+(n-1)d$

Substitute 100 for n.

$\Rightarrow a_{100}=a+(100-1)d$

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

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

$\Rightarrow a_{100}=100\sqrt{3}$

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

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

${\displaystyle \frac{1}{n}},{\displaystyle \frac{n+1}{n}},{\displaystyle \frac{2n+1}{n}}..........$

Ans: The sequence is

${\displaystyle \frac{1}{n}},{\displaystyle \frac{n+1}{n}},{\displaystyle \frac{2n+1}{n}}..........$

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

The difference between consecutive terms can be calculated as follows.

$a_2-a_1$= ${\displaystyle \frac{n+1}{n}}$- ${\displaystyle \frac{1}{n}}$ = 1 , $a_3-a_2={\displaystyle \frac{2n+1}{n}}-{\displaystyle \frac{n+1}{n}}=1$

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.

$a_n=a+(n-1)d$

Substitute 50 for n.

$\Rightarrow a_{50}=a+(50-1)d$

$\Rightarrow a_{50}={\displaystyle \frac{1}{n}}+(50-1)\left(1\right)$

$\Rightarrow a_{50}={\displaystyle \frac{1}{n}}+49$

$\Rightarrow a_{50}={\displaystyle \frac{49n+1}{n}}$

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

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.

$a_n=a+(n-1)d$

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

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

$\Rightarrow n-1={\displaystyle \frac{394}{5}}$

$\Rightarrow n={\displaystyle \frac{399}{5}}$

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:

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

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

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

The common difference can be obtained as follows,

$\Rightarrow d=a_2-a_1=9{\displaystyle \frac{1}{2}}-7{\displaystyle \frac{3}{4}}={\displaystyle \frac{19}{2}}-{\displaystyle \frac{31}{4}}={\displaystyle \frac{38-31}{4}}={\displaystyle \frac{7}{4}}$

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

$a_n=a+(n-1)d$

Substitute n = 99

$\Rightarrow a_{99}={\displaystyle \frac{31}{4}}+(99-1){\displaystyle \frac{7}{4}}$

$\Rightarrow a_{99}={\displaystyle \frac{31}{4}}+\left(98\right){\displaystyle \frac{7}{4}}$

$\Rightarrow a_{99}={\displaystyle \frac{31}{4}}+{\displaystyle \frac{98\times 7}{4}}$

$\Rightarrow a_{99}={\displaystyle \frac{31+98\times 7}{4}}$

$\Rightarrow a_{99}={\displaystyle \frac{31+98\times 7}{4}}=179{\displaystyle \frac{1}{4}}$

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

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) ${\displaystyle \frac{3}{4}},1,1{\displaystyle \frac{1}{4}}.................,3?$

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.

$a_n=a+(n-1)d$

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

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

$\Rightarrow (n-1)={\displaystyle \frac{144}{3}}$

$\Rightarrow n=48+1=49$

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.

$a_n=a+(n-1)d$

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

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

$\Rightarrow (n-1)={\displaystyle \frac{0.6}{0.03}}$

$\Rightarrow n=20+1=21$

There are 21 terms in the sequence.

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

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

The common difference can be calculated as follows.

d = 1 - ${\displaystyle \frac{3}{4}}$ = ${\displaystyle \frac{5}{4}}$ - 1

d = ${\displaystyle \frac{1}{4}}$

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

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

$a_n=a+(n-1)d$

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

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

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

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

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.

$a_n=a+(n-1)d$

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

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

$\Rightarrow n-1=17$

$\Rightarrow n=18$

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.

$a_n=a+(n-1)d$

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

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

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

The sixth  term of the A.P. is 5.

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

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

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

$\Rightarrow a_{11}=a+(11-1)d$

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

Subtract eq(1) from eq(2)-

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

$\Rightarrow -1=d$

Substitute d = -1 in eq(2) -

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

$\Rightarrow a=10$

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

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

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

$\Rightarrow a_{11}=0$

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. = $a+(n-1)d$

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

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

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

$t_2$+$t_9=17$

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

$\Rightarrow 2a+9d=17$

Substitute value of ‘a’ from eq(1):

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

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

$\Rightarrow 3d=-3$

$\Rightarrow d=-1$

Substitute d = -1 in eq(1):

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

$\Rightarrow a=13$

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.

$a_n=a+(n-1)d$

Substitute n = 10

$\Rightarrow a_{10}=254+(10-1)(-5)$

$\Rightarrow a_{10}=254-45$

$\Rightarrow a_{10}=209$

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.

$a_n=a+(n-1)d$

Given that, 

The$3^{rd}$  term of AP is 5.

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

$a_n=a+(n-1)d$

Given that, 

The${10}^{th}$  term of AP is 38.

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

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

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

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

$\Rightarrow (n-1)={\displaystyle \frac{-102}{-3}}$

$\Rightarrow n=34+1=35$

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

Consider 0 as $n^{th}$  term.

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

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

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

$\Rightarrow (n-1)={\displaystyle \frac{-21}{-3}}$

$\Rightarrow n=7+1=8$

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.

$a_n=a+(n-1)d$

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

$\Rightarrow 118=59d$

$\Rightarrow d={\displaystyle \frac{118}{59}}=2$

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

$a_n=a+(n-1)d$

$\Rightarrow a_{31}=7+(31-1)2$

$\Rightarrow a_{31}=7+60$

$\Rightarrow a_{31}=67$

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.

$a_n=a+(n-1)d$

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)-

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

$\Rightarrow 2d=5$

$\Rightarrow d={\displaystyle \frac{5}{2}}$

Substitute value of d in eq(1)-

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

$\Rightarrow a=12-{\displaystyle \frac{25}{2}}$

$\Rightarrow a=-{\displaystyle \frac{1}{2}}$

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

Second term can be calculated as follows,

$\Rightarrow a_2=a+d=-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{5}{2}}=2$

Third term can be calculated as follows,

$\Rightarrow a_3=a+2d=-{\displaystyle \frac{1}{2}}+2\times {\displaystyle \frac{5}{2}}={\displaystyle \frac{9}{2}}=4.5$

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.

$a_n=a+(n-1)d$

Given that, 

The$3^{rd}$  term of AP is 5.

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

$a_n=a+(n-1)d$

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

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

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

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

$\Rightarrow -20a=780d$

$\Rightarrow a={\displaystyle \frac{780d}{-20}}=-39d$------(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.

$a_n=a+(n-1)d$

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

$\Rightarrow {\displaystyle \frac{87}{3}}=n-1$

$\Rightarrow n=30$

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.

$a_n=a+(n-1)d$

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

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

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

$\Rightarrow (n-31)=13$

$\Rightarrow n=44$

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.

$a_n=a+(n-1)d$

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)-

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

$\Rightarrow a=8$

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

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

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

$\Rightarrow n-1=4$

$\Rightarrow n=5$

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.

$a_n=a+(n-1)d$

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

$\Rightarrow {\displaystyle \frac{783}{87}}=n-1$

$\Rightarrow n=9+1=10$

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.

$\Rightarrow A_n=a_n$

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

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

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

$\Rightarrow n-1={\displaystyle \frac{60}{5}}$

$\Rightarrow n=12+1=13$

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.

$a_n=a+(n-1)d$

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.

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

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

$\Rightarrow 2n=2$

$\Rightarrow n=1$

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 $k^2+4k+8,2k^2+3k+6$and$3k^2+4k+4$are in AP.

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

The common differences between the terms are equal.

$2k^2+3k+6-(k^2+4k+8)=3k^2+4k+4-(2k^2+3k+6)$

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

$\Rightarrow 0=2k$

$\Rightarrow k=0$

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.

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

$\Rightarrow b-a=c-b$

$\Rightarrow 2b=a+c$

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)-

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

$\Rightarrow 50d=95$

$\Rightarrow d={\displaystyle \frac{95}{50}}=1.9$

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.

$\Rightarrow a_4=3a$

$\Rightarrow a+3d=3a$

$\Rightarrow 3d=2a$

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

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

$\Rightarrow a_7=2a_3+1$

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

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

$\Rightarrow -a+2d=1$

Substituting value of a from eq(1)-

$\Rightarrow -{\displaystyle \frac{3}{2}}d+2d=1$

$\Rightarrow {\displaystyle \frac{d}{2}}=1$

$\Rightarrow d=2$

From eq(1)-

$\Rightarrow a={\displaystyle \frac{3}{2}}\times 2=3$

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.

$\Rightarrow a_{15}=2a_8-1$

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

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

$\Rightarrow a=2a-1$

$\Rightarrow a=1$

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

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

$\Rightarrow 7d=28$