NCERT Exemplar for Class 10 Maths Chapter 5 - Arithmetic Progressions - Free PDF Download

Multiple Choice Questions

Choose the correct answer from the given four options:

Sample Question 1 : The 10th term of the AP: 5, 8, 11, 14, ... is

(A) 32 (B) 35 (C) 38 (D) 185

Ans: Given AP is 5, 8, 11, 14,... Here, $$a=5$$ and $$d=3$$

Now, since

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

So we have

$$\Rightarrow a_{10}=5+\left(10-1\right)\left(3\right)$$

Hence,

$$\Rightarrow a_{10}=32$$

So the correct answer is (A).

Sample Question 2 : In an AP if a = –7.2, d = 3.6, $$a_n=7.2$$ , then n is

(A) 1 (B) 3 (C) 4 (D) 5

Ans: Given that, in an AP we have $$a=-7.2$$ and $$d=3.6$$

Now if $$a_n=7.2$$ then we have

$$\Rightarrow 7.2=-7.2+\left(n-1\right)\left(3.6\right)$$

That gives us,

$$\Rightarrow \left(n-1\right)={\displaystyle \frac{2\times 7.2}{3.6}}$$

i.e.

$$\Rightarrow n=5$$

Hence, the correct answer is (D).

EXERCISE 5.1

Choose the correct answer from the given four options in the following questions:

1. In an A.P., if d = –4, n = 7, $$a_n=4$$ , then a is

(a) 6

(b) 7

(c) 20

(d) 28

Ans: Since $$a_n=a+\left(n-1\right)d$$

Hence, that gives us,

$$\Rightarrow 4=a+\left(7-1\right)\left(-4\right)$$

i.e.

$$\Rightarrow a=28$$

Hence, the correct answer is (d).

2. In an A.P., if a = 3.5, d = 0, n = 101, then $$a_n$$ will be

(a) 0

(b) 3.5

(c) 103.5

(d) 104.5

Ans: Since $$a_n=a+\left(n-1\right)d$$

Hence, that gives us,

$$\Rightarrow a_n=3.5+\left(101-1\right)\left(0\right)$$

i.e.

$$\Rightarrow a_n=3.5$$

Hence, the correct answer is (b).

3. The list of numbers  –10, –6, –2, 2, … is

(a) an A.P. with d = –16

(b) an A.P. with d = 4

(c) an A.P. with d = –4

(d) not an A.P.

Ans: The given list of numbers  –10, –6, –2, 2, … is an A.P. where $$a=-10$$ and $$d=4$$

Hence, the correct answer is (b).

4. The 11th term of the A.P. $$-5,{\displaystyle \frac{-5}{2}},0,{\displaystyle \frac{5}{2}},...$$ is

(a) –20

(b) 20

(c) –30

(d) 30

Ans: In the given sequence, we have $$a=-5$$ and $$d={\displaystyle \frac{5}{2}}$$

Hence, $$a_{11}$$ will be given by,

$$\Rightarrow a_{11}=-5+\left(11-1\right){\displaystyle \frac{5}{2}}$$

i..e

$$\Rightarrow a_{11}=20$$

Hence, the correct answer is (b).

5. The first four terms of an A.P., whose first term is –2 and the common difference is (– 2), are

(a) –2, 0, 2, 4

(b) –2, 4, –8, 16

(c) –2, –4, –6, –8

(d) –2, –4, –8, –16

Ans: Given that $$a=-2$$ and $$d=-2$$

Therefore, the given A.P is $$-2,-4,-6,-8,-10,...$$

Hence, the correct answer is (c).

6. The 21st term of the A.P. whose first two terms are –3 and 4 is

(a) 17

(b) 137

(c) 143

(d) –143

Ans: Given that $$a_1=-3$$ and $$a_2=4$$ hence, $$d=a_2-a_1=4-\left(-3\right)$$ i.e. $$d=7$$

Hence,

$$\Rightarrow a_{21}=-3+\left(21-1\right)7$$

i.e.

$$\Rightarrow a_{21}=137$$

Hence, the correct answer is (b).

7. If the 2nd term of an A.P. is 13 and 5th term is 25, what is its 7th term ?

(a) 30

(b) 33

(c) 37

(d) 38

Ans: Given that $$a_2=a+d=13$$ and $$a_5=a+4d=25$$ hence we have

$$\Rightarrow a+4d-a-d=25-13=12$$

That gives

$$\Rightarrow 3d=12$$

i.e.

$$\Rightarrow d=4$$

So $$a_1=13-4=9$$

And hence,

$$\Rightarrow a_7=9+\left(7-1\right)4$$

i.e.

$$\Rightarrow a_7=33$$

Hence, the correct answer is (b).

8. Which term of the A.P. : 21, 42, 63, 84, … is 210 ?

(a) 9th

(b) 10th

(c) 11th

(d) 12th

Ans: Given that $$a=21$$ , $$a_n=210$$ and $$d=42-21=21$$

Hence,

$$\Rightarrow 210=21+\left(n-1\right)21$$

i.e.

$$\Rightarrow n=10$$

Hence, the correct answer is (c).

9. If the common difference of an A.P. is 5, then what is $$a_{18}-a_{13}$$ ?

(a) 5

(b) 20

(c) 25

(d) 30

Ans: Given that, $$d=5$$

Now, $$a_{18}=a+17d$$ and $$a_{13}=a+12d$$

Hence,

$$\Rightarrow a_{18}-a_{13}=a+17d-a-12d$$

i.e.

$$\Rightarrow a_{18}-a_{13}=5d$$

That is,

$$\Rightarrow a_{18}-a_{13}=25$$

Hence, the correct answer is (c).

10. What is the common difference of an A.P. in which $$a_{18}-a_{14}=32$$ ?

(a) 8

(b) –8

(c) –4

(d) 4

Ans: Given that, $$a_{18}-a_{14}=32$$ i.e. $$a+17d-a-13d=32$$

Hence we have

$$\Rightarrow a+17d-a-13d=32$$

Or,

$$\Rightarrow 4d=32$$

So the common difference is $$d=8$$

Hence, the correct answer is (a).

11. Two APs have the same common difference. The 1st term of one of these is –1 and that of the other is –8. Then the difference between their 4th terms is

(a) –1

(b) –8

(c) 7

(d) –9

Ans: Two APs have the same common difference, say d.

The fourth term of first AP will be $$-1+3d$$ and that of the second AP will be $$-8+3d$$

Hence, their difference is $$-1+3d-\left(-8+3d\right)=7$$

Hence, the correct answer is (c).

12. If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, then its 18th term will be

(a) 7 

(b) 11

(c) 18

(d) 0

Ans: According to the question, we have

$$\Rightarrow 7\left(a+6d\right)=11\left(a+10d\right)$$

That gives

$$\Rightarrow 4a+68d=0$$

i.e.

$$\Rightarrow 4\left(a+17d\right)=0$$

Or,

$$\Rightarrow a+17d=0$$

Hence, the correct answer is (d).

13. The 4th term from the end of the A.P. –11, –8, –5, …, 49 is

(a) 37

(b) 40

(c) 43

(d) 58

Ans: Given AP is –11, –8, –5, …, 49

Consider the reverse of this AP, i.e. where $$a=49$$ and $$d=-3$$

Now, we have

$$\Rightarrow a_4=a+3d$$

i.e.

$$\Rightarrow a_4=49+3\left(-3\right)$$

Or,

$$\Rightarrow a_4=40$$

Hence, the correct answer is (b).

14. The famous mathematician associated with finding the sum of the first 100 natural numbers is

(a) Pythagoras

(b) Newton

(c) Gauss

(d) Euclid

Ans: Gauss was the first mathematician associated with finding the sum of the first 100 natural numbers.

The sum of first natural numbers can be found considering the AP 1,2,3,...100

Here, we have $$a=d=1$$ and $$n=100$$

The sum of an AP is given by the formula

$$\Rightarrow S_n={\displaystyle \frac{n}{2}}\left[2a+\left(n-1\right)d\right]$$

That gives us,

$$\Rightarrow S_{100}={\displaystyle \frac{100}{2}}\left[2\left(1\right)+\left(100-1\right)\cdot 1\right]$$

Or,

$$\Rightarrow S_{100}=50\left[2+99\right]$$

i.e.

$$\Rightarrow S_{100}=5050$$

Hence, the correct answer is (c).

15. If the first term of an A.P. is –5 and the common difference is 2, then the sum of first 6 terms is

(a) 0

(b) 5

(c) 6

(d) 15

Ans: Given that $$a=-5$$ and $$d=2$$ and $$n=6$$

Hence,

$$\Rightarrow S_6={\displaystyle \frac{6}{2}}\left[2\left(-5\right)+\left(6-1\right)\cdot 2\right]$$

Or,

$$\Rightarrow S_6=3\left[-10+10\right]$$

i.e.

$$\Rightarrow S_6=0$$

Hence, the correct answer is (a).

16. The sum of first 16 terms of the A.P. 10, 6, 2, … is

(a) –320

(b) 320

(c) –352

(d) –400

Ans: Given that $$a=10$$ and $$d=6-10=-4$$ and $$n=16$$

Hence,

$$\Rightarrow S_{16}={\displaystyle \frac{16}{2}}\left[2\left(10\right)+\left(16-1\right)\cdot \left(-4\right)\right]$$

Or,

$$\Rightarrow S_{16}=8\left[20-60\right]$$

i.e.

$$\Rightarrow S_{16}=-320$$

Hence, the correct answer is (a).

17. In an A.P., if a = 1, $$a_n=20$$ and $$S_n=399$$ then n is

(a) 19

(b) 21

(c) 38

(d) 42

Ans: Given that, $$a=1$$ , $$a_n=20$$ and $$S_n=399$$ 

Therefore, we can write

$$\Rightarrow S_n={\displaystyle \frac{n}{2}}\left[2a+\left(n-1\right)\cdot d\right]$$

i.e.

$$\Rightarrow S_n={\displaystyle \frac{n}{2}}\left[a+a_n\right]$$

Or,

$$\Rightarrow 399={\displaystyle \frac{n}{2}}\left[1+20\right]$$

That gives,

$$\Rightarrow n={\displaystyle \frac{798}{21}}=38$$

Hence, the correct answer is (c).

18. The sum of first five multiples of 3 is

(a) 45

(b) 55

(c) 65

(d) 75

Ans: The sum of first five multiples of 3 can be found considering $$a=3$$ , $$d=3$$ and $$n=5$$

Hence,

$$\Rightarrow S_5={\displaystyle \frac{5}{2}}\left[2\left(3\right)+\left(5-1\right)\cdot 3\right]$$

i.e.

$$\Rightarrow S_5={\displaystyle \frac{5}{2}}\left[18\right]$$

Or,

$$\Rightarrow S_5=45$$

Hence, the correct answer is (a).

Short Answer Questions with Reasoning

Sample Question 1: In the AP: 10, 5, 0, –5, ... the common difference d is equal to 5. Justify whether the above statement is true or false.

Ans: Given that,  an AP as 10,5,0,-5,...

Here we have the first term, i.e. $$a=10$$ and the common difference as $$d=5-10$$

Hence, we have $$d=-5$$

So the given statement is false that the common difference d is equal to 5 because it is -5.

Sample Question 2 : Divya deposited Rs 1000 at compound interest at the rate of 10% per annum. The amounts at the end of first year, second year, third year, ..., form an AP. Justify your answer.

Ans: Given that, Divya deposited Rs 1000 at compound interest at the rate of 10% per annum.

Here, the amounts at the end of each year can be given as $$1100,1210,1331,...$$

Since the common difference in the above sequence is not unique and varies between each interval, therefore the given sequence does not form an AP.

Sample Question 3: The nth term of an AP cannot be $$n^2+1$$ . Justify your answer.

Ans: Consider, $$a_n=n^2+1$$

Then we can write,

$$\Rightarrow a_1={\left(1\right)}^2+1=2$$

And,

$$\Rightarrow a_2={\left(2\right)}^2+1=5$$

Also,

$$\Rightarrow a_3={\left(3\right)}^2+1=10$$

Here, $$a_2-a_1=3$$ but $$a_3-a_2=5$$

Since the common difference is not unique and varies between each consecutive terms, hence $$a_n=n^2+1$$ can be the nth term of an AP.

EXERCISE 5.2

1. Which of the following form an A.P. ? Justify your answer.

(i) –1, –1, –1, –1, …

Ans: Here, all the terms are equal to each other i.e. the common difference is zero everywhere. So, the sequence –1, –1, –1, –1,… forms an AP.

(ii) 0, 2, 0, 2, …

Ans: Here, the given sequence is 0, 2, 0, 2,… where the common difference is not unique and varying between 2 and -2. Hence, the sequence 0, 2, 0, 2,… does not form an AP.

(iii) 1, 1, 2, 2, 3, 3, …

Ans: Here, the given sequence is 1, 1, 2, 2, 3, 3,… where the common difference is not unique and varying between 0 and 1. Hence, the sequence 1, 1, 2, 2, 3, 3,… does not form an AP.

(iv) 11, 22, 33, …

Ans: Here, the given sequence is 11, 22, 33,… where the common difference is unique and constant everywhere i.e. 11 . Hence, the sequence 11, 22, 33,… forms an AP.

(v) $${\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{4}},...$$

Ans: Here, the given sequence is $${\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{4}},...$$ where the common difference is not unique and varying everywhere. We have $${\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{2}}\ne {\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{3}}$$ i.e. $$-{\displaystyle \frac{1}{6}}\ne -{\displaystyle \frac{1}{12}}$$

Hence, the sequence $${\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{4}},...$$ does not form an AP.

(vi) $$2,2^2,2^3,2^4,...$$

Ans: Here, the given sequence is $$2,2^2,2^3,2^4,...$$ where the common difference is not unique and varying everywhere. We have $$4-2\ne 8-4\ne 16-8$$ i.e. $$2\ne 4\ne 8$$

Hence, the sequence $$2,2^2,2^3,2^4,...$$ does not form an AP.

(vii) $$\sqrt{3},\sqrt{12},\sqrt{27},\sqrt{48},...$$

Ans: Here, the given sequence is $$\sqrt{3},\sqrt{12},\sqrt{27},\sqrt{48},...$$ where the common difference is not unique and varies everywhere. We have $$\sqrt{12}-\sqrt{3}\ne \sqrt{27}-\sqrt{12}\ne \sqrt{48}-\sqrt{27}$$

Hence, the sequence $$\sqrt{3},\sqrt{12},\sqrt{27},\sqrt{48},...$$ does not form an AP.

2. Justify whether it is true to say that $$-1,{\displaystyle \frac{-3}{2}},-2,{\displaystyle \frac{5}{2}},...$$ forms an A.P. as $$a_2-a_1=a_3-a_2$$ .

Ans: Given sequence is $$-1,{\displaystyle \frac{-3}{2}},-2,{\displaystyle \frac{5}{2}},...$$

Here, we have $${\displaystyle \frac{-3}{2}}-\left(-1\right)={\displaystyle \frac{-1}{2}}$$ and $$-2-\left({\displaystyle \frac{-3}{2}}\right).={\displaystyle \frac{-1}{2}}$$

But, $${\displaystyle \frac{5}{2}}-\left(-2\right)={\displaystyle \frac{9}{2}}$$

Hence, the common difference is not constant everywhere so the given sequence is not an AP.

3. For the A.P. –3, –7, –11, …, can we find directly $$a_{30}-a_{20}$$ without actually finding $$a_{30}$$ and $$a_{20}$$ ? Give reasons for your answer.

Ans: Given AP is –3, –7, –11,... Here $$d=-7-\left(-3\right)=-4$$

Now,

$$\Rightarrow a_{30}-a_{20}=a+29d-\left(a+19d\right)$$

That is,

$$\Rightarrow a_{30}-a_{20}=10d$$

Hence,

$$\Rightarrow a_{30}-a_{20}=-40$$

4. Two A.P.s have the same common difference. The first term of one A.P. is 2, and that of the other is 7. The difference between their 10th terms is same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why ?

Ans: Given that, two A.P.s have the same common difference, say d.

Now their first terms are $$a=2$$ and $$b=7$$ respectively.

The difference between their 10th terms is,

$$\Rightarrow a_{10}-b_{10}=a+9d-\left(b+9d\right)$$

i.e.

$$\Rightarrow a_{10}-b_{10}=a-b=-5$$

Also, the difference between their 21st terms is,

$$\Rightarrow a_{21}-b_{21}=a+20d-\left(b+20d\right)$$

i.e.

$$\Rightarrow a_{21}-b_{21}=a-b=-5$$

Now, the difference between any two corresponding terms is,

$$\Rightarrow a_n-b_n=a+\left(n-1\right)d-\left[b+\left(n-1\right)d\right]$$

i.e.

$$\Rightarrow a_n-b_n=a-b=-5$$

Therefore, the difference between their 10th terms is same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms.

5. Is 0 a term of the A.P. 31, 28, 25, … ∵ Justify your answer.

Ans: In the given AP, we have $$a=31$$ , $$d=28-31=-3$$

Now if $$a_n=0$$ , then we have

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

That gives

$$\Rightarrow 0=31+\left(n-1\right)\left(-3\right)$$

Or,

$$\Rightarrow \left(n-1\right)=-{\displaystyle \frac{31}{\left(-3\right)}}$$

i.e.

$$\Rightarrow n={\displaystyle \frac{31+3}{3}}$$

Hence,

$$\Rightarrow n={\displaystyle \frac{34}{3}}\notin \mathbb{N}$$

Therefore, 0 is not a term of the given AP.

6.The taxi fare after each km, when the fare is Rs 15 for the first km and Rs 8 for each additional km, does not form an A.P., as the total fare (in Rs) after each km is 15, 8, 8, 8, … Is the statement true ∵ Give reasons.

Ans: Yes, the sequence of the total fare does not form an AP. This statement is true because the total fare that is given as 15, 8, 8, 8, … is not the total fare of 1,2,3,4,5,.. kilometres respectively.

Here, we have $$a=15$$  and $$d_1=8-15=-7$$

But, $$d_1=-7\ne d_2=d_3=d_4=...=d_n=0$$

The given sequence is in the form of $$a,d,d,d,d...$$ instead of $$a,a+d,a+2d,a+3d,...$$

Hence, the given terms do not form an AP.

7. In which of the following situations do the lists of numbers involved form an A.P.? Give reasons for your answers.

(i) The fee charged from a student every month by a school for the whole session, when the monthly fee is Rs 400.

(ii) The fee charged every month by a school from classes I to XII, when the monthly fee for class I is Rs 250, and it increase by Rs 50 for the next higher class.

(iii) The amount of money in the account of Varun at the end of every year when Rs 1000 is deposited at simple interest of 10% per annum.

(iv) The number of bacteria in a certain food item after each second, when they double in every second.

Ans: (i) If the monthly fee is Rs 400, then the fee charged from a student every month by a school for the whole session, is given by $$400,400,400,....$$

Here, $$a=400$$ and $$d=0$$ .

Therefore, it forms an AP.

(ii) If the monthly fee for class I is Rs 250, and it increases by Rs 50 for the next higher class.

Then the fee charged every month by a school from classes I to XII, is given by 250,300,350,400,...

Here, $$a=250$$ and $$d=50$$ .

Therefore, it forms an AP.

(iii) If Rs 1000 is deposited at simple interest of 10% per annum, then the amount of money in the account of Varun at the end of every year is given by, 1000,1100,1200,1300,...

Here $$a=1000$$ and $$d=100$$

Therefore, it does not form an AP.

(iv) If the number of bacteria double in every second in a certain food item, then the number of bacteria after each second is given by a,2a,4a,8a,16a,...

Here the common difference is not unique and varies at every interval.

Therefore, it does not form an AP.

8. Justify whether it is true to say that the following are the nth terms of an A.P.

(i) $$2n–3$$

Ans: We have $$a_n=2n-3$$

Now, we can write $$a_1=2-3=-1$$ , $$a_2=4-3=1$$ , $$a_3=6-3=3$$ and so on.

Here, we have $$a=-1$$ and $$d=2$$

Hence 2n – 3 is an nth term of an A.P.

(ii) $$3n^2+5$$ 

Ans: We have $$a_n=3n^2+5$$

Now, we can write $$a_1=3+5=8$$ , $$a_2=12+5=17$$ , $$a_3=27+5=32$$ and so on.

Here the common difference is not unique and varies at every interval.

Hence $$3n^2+5$$ is not an nth term of an A.P.

(iii) $$1+n+n^2$$

Ans: We have $$a_n=1+n+n^2$$

Now, we can write $$a_1=1+1+1=3$$ , $$a_2=1+2+4=7$$ , $$a_3=1+3+9=13$$ and so on.

Here the common difference is not unique and varies at every interval.

Hence $$1+n+n^2$$ is not an nth term of an A.P.

Short Answer Questions

Sample Question 1 : If the numbers n – 2, 4n – 1 and 5n + 2 are in AP, find the value of n.

Ans: Given that, three numbers $$n-2$$ , $$4n-1$$ and $$5n+2$$ are in AP.

Now we have

$$\Rightarrow \left(4n-1\right)-\left(n-2\right)=3n+1$$

And,

$$\Rightarrow \left(5n+2\right)-\left(4n-1\right)=n+3$$

Now if they are if then the common difference must be constant.

Hence, we can write

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

That gives us,

$$\Rightarrow 2n=2$$

Therefore,

$$\Rightarrow n=1$$

Sample Question 2 : Find the value of the middle most term (s) of the AP : –11, –7, –3,..., 49.

Ans: Given AP is –11, –7, –3,..., 49.

Here we have $$a=-11$$ , $$d=4$$ and $$a_n=49$$

So we can write,

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

Or,

$$\Rightarrow 49=-11+\left(n-1\right)\left(4\right)$$

That gives us,

$$\Rightarrow n=16$$

Since n is an even number, so there are two middle terms i.e. the $${\displaystyle \frac{n}{2}}th$$ and $$\left({\displaystyle \frac{n}{2}}+1\right)th$$ terms.

Hence the middle most terms are $$a_8$$ and $$a_9$$ .

Now,

$$\Rightarrow a_8=-11+7\left(4\right)$$

i.e.

$$\Rightarrow a_8=17$$

Ans,

$$\Rightarrow a_9=-11+8\left(4\right)$$

i.e.

$$\Rightarrow a_9=21$$

Sample Question 3: The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, find the AP.

Ans: According to the given question, we have $$a_1+a_2+a_3=33$$ 

Let $$a_1=a-d$$ , $$a_2=a$$ and $$a_3=a+d$$

Now that can be written as,

$$\Rightarrow a-d+a+a+d=33$$

i.e.

$$\Rightarrow 3a=33$$

Hence,

$$\Rightarrow a=11$$

Now it is also given that $$a_1{a}_3=a_2+29$$

That gives us,

$$\Rightarrow \left(a-d\right)\left(a+d\right)=a+29$$

That is,

$$\Rightarrow \left(11-d\right)\left(11+d\right)=11+29$$

Or,

$$\Rightarrow {\left(11\right)}^2-d^2=40$$

That can be written as,

$$\Rightarrow d^2=121-40$$

Hence,

$$\Rightarrow d^2=81$$

i.e.

$$\Rightarrow d=\pm 9$$

Therefore, the required AP is either $$2,11,20,29,38,...$$ or $$20,11,2,-7,...$$

EXERCISE 5.3

1. Match the A.P.s given in column A with suitable common differences given in column B.

Ans: (1) We have 2, –2, –6, –10,...

Here, we have

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

(2) We have a = –18, n = 10, $$a_n=0$$

Now, we can write

$$\Rightarrow 0=-18+\left(10-1\right)d$$

Hence,

$$\Rightarrow d=2$$

(3) We have a = 0, $$a_{10}=6$$

Hence, we can write

$$\Rightarrow 6=0+\left(10-1\right)d$$

i.e.

$$\Rightarrow d={\displaystyle \frac{2}{3}}$$

(4) We have$$a_2=13$$ , $$a_4=3$$

Hence, we can write

$$\Rightarrow a+3d-\left(a+d\right)=13-3$$

Or,

$$\Rightarrow 2d=10$$

i.e.

$$\Rightarrow d=5$$

2. Verify that each of the following is an A.P. and then write its next three terms.

(i) $$0,{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{4}},...$$

Ans: Here, we have $$a=0$$ and $$d_1={\displaystyle \frac{1}{4}}-0={\displaystyle \frac{1}{4}}$$ , $$d_2={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{4}}={\displaystyle \frac{1}{4}}$$ , $$d_3={\displaystyle \frac{3}{4}}-{\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{4}}$$

Therefore, $$d_{}={\displaystyle \frac{1}{4}}$$ so the given terms form an AP.

Its next three terms will be given by $$0,{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{4}},1,{\displaystyle \frac{5}{4}},{\displaystyle \frac{3}{2}},{\displaystyle \frac{7}{4}},{\displaystyle \frac{8}{4}},...$$ 

(ii) $$5,{\displaystyle \frac{14}{3}},{\displaystyle \frac{13}{3}},4,...$$

Ans: Here, we have $$a=5$$ and $$d_1={\displaystyle \frac{14}{3}}-5=-{\displaystyle \frac{1}{3}}$$ , $$d_2={\displaystyle \frac{13}{3}}-{\displaystyle \frac{14}{3}}=-{\displaystyle \frac{1}{3}}$$ , $$d_3=4-{\displaystyle \frac{13}{3}}=-{\displaystyle \frac{1}{3}}$$

Therefore, $$d=-{\displaystyle \frac{1}{3}}$$ so the given terms form an AP.

Its next three terms will be given by $$5,{\displaystyle \frac{14}{3}},{\displaystyle \frac{13}{3}},4,{\displaystyle \frac{11}{3}},{\displaystyle \frac{10}{3}},3,{\displaystyle \frac{8}{3}},...$$

(iii) $$\sqrt{3},2\sqrt{3},3\sqrt{3},...$$

Ans: Here, we have $$a=\sqrt{3}$$ and $$d_1=2\sqrt{3}-\sqrt{3}=\sqrt{3}$$ , $$d_2=3\sqrt{3}-2\sqrt{3}=\sqrt{3}$$ , $$d_3=4\sqrt{3}-3\sqrt{3}=\sqrt{3}$$

Therefore, $$d=\sqrt{3}$$ so the given terms form an AP.

Its next three terms will be given by $$\sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3},5\sqrt{3},6\sqrt{3},7\sqrt{3},...$$

(iv) a + b, (a + 1) + b, (a + 1) + (b + 1), …

Ans: Here, we have $$a_1=a+b$$ and $$d_1=\left(a+1\right)+b-a-b=1$$ , $$d_2=\left(a+1\right)+\left(b+1\right)-\left(a+1\right)-b=1$$ 

Therefore, $$d=1$$ so the given terms form an AP.

Its next three terms will be given by $$\left(a+b\right),\left(a+1\right)+b,\left(a+1\right)+\left(b+1\right),\left(a+2\right)+\left(b+1\right),\left(a+2\right)+\left(b+2\right),\left(a+3\right)+\left(b+1\right),...$$

(v) a, 2a + 1, 3a + 2, 4a + 3, …

Ans: Here, we have $$a_1=a$$ and $$d_1=\left(2a+1\right)-a=a+1$$ , $$d_2=\left(3a+2\right)-\left(2a+1\right)=a+1$$ , $$d_3=\left(4a+3\right)-\left(3a+2\right)=a+1$$

Therefore, $$d=a+1$$ so the given terms form an AP.

Its next three terms will be given by $$a,\left(2a+1\right),\left(3a+2\right),\left(4a+3\right),\left(5a+4\right),\left(6a+5\right),\left(7a+6\right),...$$

3. Write the first three terms of the A.P.s when a and d are as given below.

(i) $$a={\displaystyle \frac{1}{2}},d={\displaystyle \frac{-1}{6}}$$

Ans: Here, $$a={\displaystyle \frac{1}{2}},d={\displaystyle \frac{-1}{6}}$$

Hence, the AP is given by $${\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{6}},{\displaystyle \frac{1}{2}}-{\displaystyle \frac{2}{6}},{\displaystyle \frac{1}{2}}-{\displaystyle \frac{3}{6}},{\displaystyle \frac{1}{2}}-{\displaystyle \frac{4}{6}},..$$

That is, $${\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{6}},0,{\displaystyle \frac{-1}{6}},..$$

(ii) $$a=-5,d=-3$$

Ans: Here, $$a=-5,d=-3$$

Hence, the AP is given by $$-5,-5-3,-5-6,-5-9,-5-12,...$$

That is, $$-5,-8,-11,-14,-17,...$$

(iii) $$a=\sqrt{2},d={\displaystyle \frac{1}{\sqrt{2}}}$$

Ans: Here, we have $$a=\sqrt{2}$$ and $$d={\displaystyle \frac{1}{\sqrt{2}}}$$

Hence, the terms of this AP are given as 

$$\Rightarrow a,a+d,a+2d,a+3d,...$$

That gives us,

$$\Rightarrow \sqrt{2},\sqrt{2}+{\displaystyle \frac{1}{\sqrt{2}}},\sqrt{2}+{\displaystyle \frac{2}{\sqrt{2}}},\sqrt{2}+{\displaystyle \frac{3}{\sqrt{2}}},...$$

Or,

$$\Rightarrow \sqrt{2},{\displaystyle \frac{3\sqrt{2}}{2}},2\sqrt{2},{\displaystyle \frac{5}{\sqrt{2}}},3\sqrt{2,}...$$

Therefore, the given AP is $$\sqrt{2},{\displaystyle \frac{3\sqrt{2}}{2}},2\sqrt{2},{\displaystyle \frac{5}{\sqrt{2}}},3\sqrt{2,}...$$

4. Find a, b and c such that the following numbers are in A.P.: a, 7, b, 23, c.

Ans: Given terms are, $$a,7,b,23,c,...$$

For the given terms to from an AP, we must have $$7-a=b-7=23-b=c-23$$

That gives us, $$a+b=14,2b=30,b+c=46$$

Hence, $$b=15$$ so $$a=-1$$ and $$c=31$$

So the given AP is $$-1,7,15,23,31,...$$

5. Determine the A.P. whose 5th term is 19 and the difference of the 8th term from the 13th term is 20.

Ans: Given that $$a_5=19$$ and $$a_{13}-a_8=20$$

That gives us, $$a+4d=19$$ and $$a+12d-a-7d=20$$or $$5d=20$$

Hence, $$d=4$$

So we have $$a=19-4\left(4\right)=3$$

Hence, the given AP is $$3,7,11,15,19,...$$

6. The 26th, 11th and the last term of an A.P. are 0, 3 and $$-{\displaystyle \frac{1}{5}}$$ respectively. Find the common difference and the number of terms.

Ans: Given that $$a_{26}=0,a_{11}=3,a_n=-{\displaystyle \frac{1}{5}}$$

That gives us $$a_{26}-a_{11}=a+25d-a-10d=0-3$$ i.e. $$15d=-3$$

Hence, $$d=-{\displaystyle \frac{1}{5}}$$ and $$a+25\left(-{\displaystyle \frac{1}{5}}\right)=0$$ i.e. $$a=5$$

Now if $$-{\displaystyle \frac{1}{5}}=5+\left(n-1\right)\left(-{\displaystyle \frac{1}{5}}\right)$$

Then we have

$$\Rightarrow 26=\left(n-1\right)$$

i.e.

$$\Rightarrow n=27$$

7. The sum of the 5th and the 7th terms of an A.P. is 52, and the 10th term is 46. Find the A.P.

Ans: Given that $$a_{10}=46$$ and $$a_5+a_7=52$$

That gives us $$a_5+a_7=a+4d+a+6d=52$$ i.e. $$2a+10d=52$$

Hence $$a+5d=26$$

Also we have $$a_{10}=a+9d=46$$

Therefore, $$a+9d-a-5d=46-26$$

Hence, $$4d=20$$ i.e. $$d=5$$

That gives us, $$a+9\left(5\right)=46$$ i.e. $$a=1$$

Therefore, the AP is $$1,6,11,16,21,...$$

8. Find the 20th term of an A.P. whose 7th term is 24 less than the 11th term, first term being 12.

Ans: Given that $$a=12$$ and $$a_7=a_{11}-24$$

That gives us,

$$\Rightarrow a_{11}-a_7=a+10d-a-6d=24$$

Or,

$$\Rightarrow 4d=24$$

Hence,

$$\Rightarrow d=6$$

Now, the 20th term is given by

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

i.e.

$$\Rightarrow a_{20}=12+\left(20-1\right)\cdot \left(6\right)$$

That is

$$\Rightarrow a_{20}=126$$

9. If the 9th term of an A.P. is zero, prove that its 29th term is twice its 19th term.

Ans: Given that $$a_9=0$$ i.e. $$a+8d=0$$ that gives us $$a=-8d$$

Now we can write $$a_{29}=a+28d$$ i.e. $$a_{29}=-8d+28d$$

Hence, $$a_{29}=20d$$

Also, we have $$a_{19}=a+18d$$ i.e. $$a_{19}=-8d+18d$$

Hence, $$a_{19}=10d$$

Multiplying both sides by 2, we get

$$\Rightarrow 2a_{19}=20d$$

That gives us,

$$\Rightarrow a_{29}=2a_{19}$$

10. Find whether 55 is a term of the A.P.: 7, 10, 13, … or not. If yes, find which term it is.

Ans: Given terms are 7,10,13,...

Here $$a=7$$ and $$d=3$$

Now if $$a_n=55$$ then we can write,\

$$\Rightarrow 55=7+\left(n-1\right)3$$

That gives,

$$\Rightarrow \left(n-1\right)={\displaystyle \frac{48}{3}}$$

i.e.

$$\Rightarrow n=17$$

Hence 55 is the 17th term of the AP 7,10,13,...

11. Determine k so that $$\left(k^2+4k+8\right)$$  , $$\left(2k^2+3k+6\right)$$ and $$\left(3k^2+4k+4\right)$$ are three consecutive terms of an A.P.

Ans: The three given terms will be consecutive terms of an AP if the common difference among them is constant.

Here, we have

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

And,

$$\Rightarrow \left(3k^2+4k+4\right)-\left(2k^2+3k+6\right)=k^2+k-2$$

Now if the above two differences are equal then,

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

That gives us,

$$\Rightarrow -k=k$$

Hence,

$$\Rightarrow k=0$$

12. Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.

Ans: Let the three consecutive terms of an AP be $$a-d,a,a+d$$

Now according to the question, their sum is 207.

Hence,

$$\Rightarrow \left(a-d\right)+a+\left(a+d\right)=207$$

That gives us,

$$\Rightarrow 3a=207$$

Hence,

$$\Rightarrow a=69$$

Also, the sum of two smaller parts is 4623.

Hence,

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

That gives us,

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

Or,

$$\Rightarrow 69-d={\displaystyle \frac{4623}{69}}$$

i.e

$$\Rightarrow 69-d=67$$

Hence,

$$\Rightarrow d=2$$

Hence, the three numbers are $$67,69,71$$ .

13. The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles of the triangle.

Ans: The angles of a triangle are given to be in an A.P. Let these angles be $${\left(a-d\right)}^{\circ },a^{\circ },{\left(a+d\right)}^{\circ }$$ respectively.

Now by the angle sum property of a triangle, we have

$$\Rightarrow {\left(a-d\right)}^{\circ }+a^{\circ }+{\left(a+d\right)}^{\circ }={180}^{\circ }$$

That gives us,

$$\Rightarrow {\left(3a\right)}^{\circ }={180}^{\circ }$$

i.e.

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

Also, the greatest angle is twice the least, so we have

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

That gives us,

$$\Rightarrow {\left(60+d\right)}^{\circ }=2{\left(60-d\right)}^{\circ }$$

Or,

$$\Rightarrow 3d^{\circ }={\left(120-60\right)}^{\circ }$$

i.e.

$$\Rightarrow d^{\circ }={20}^{\circ }$$

Hence, the angles of the triangle are $${40}^{\circ },{60}^{\circ },{80}^{\circ }$$ .

14. If nth terms of two A.P.s: 9, 7, 5, … and 24, 21, 18, … are same, then find the values of n. Also find that term.

Ans: Given that, two APs that are 9, 7, 5, … where $$a=9,d_1=-2$$

And 24, 21, 18, … where $$b=24,d_2=-3$$

According to the question, we have $$a_n=b_n$$

That gives us,

$$\Rightarrow a+\left(n-1\right)d_1=b+\left(n-1\right)d_2$$

Or,

$$\Rightarrow 9+\left(n-1\right)\cdot \left(-2\right)=24+\left(n-1\right)\cdot \left(-3\right)$$

That is,

$$\Rightarrow 11-2n=27-3n$$

Hence,

$$\Rightarrow n=16$$

Now,

$$\Rightarrow a_{16}=9+15\left(-2\right)$$

Hence,

$$\Rightarrow a_{16}=-21$$