# Arithmetic ProgressionsExercise – 5.2

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the A.P:

 a d n an (i) 7 3 8 … (ii) -18 … 10 0 (iii) … -3 18 -5 (iv) -18.9 2.5 … 3.6 (v) 3.5 0 105 …

(i) Here...9Here aksjdfksjafj, a = 7, d = 3 and n = 8 Therefore, substituting the value in an = a + (n – 1)d,

an = 7 + (8 – 1)3

an = 7 + 21 = 28

(ii) Here...9HereHe, a = -18, n = 10 and an = 0 Therefore, substituting the value in an = a + (n – 1)d,

0 = -18 + (10 – 1)d

18 = 9d

d = 2

(iii) Here...9HE, d = -3, n = 18 and an = -5 Therefore, substituting the value in an = a + (n – 1)d,

-5 = a + (18 – 1)×-3

-5 = a + 17×-3

-5 = a – 51

a = 51 – 5 = 46

(iv) Here...9, a = -18.9, d = 2.5 and an = 3.6 Therefore, substituting the value in an = a + (n – 1)d,

3.6 = -18.9 + (n – 1)(2.5)

3.6 = -18.9 + 2.5n – 2.5

2.5n = 3.6 + 21.4

2.5n  = 25.0

n = 10

(v) Here...9, a = 3.5, d = 0 and n = 105 Therefore, substituting the value in an = a + (n – 1)d,

an = 3.5 + (105 – 1)0

an = 3.5 + 0 = 3.5

2. Choose the correct choice in the following and justify:

(i) 30th term of the A.P.: 10, 7, 4, ……, is:

(A) 97                  (B) 77                    (C) -77               (D) -87

Here a = 10, d = 7 – 10 = - 3 and n = 30,

Therefore, substituting the value in an = a + (n – 1)d,

a30 = 10 + (30 – 1)×-3

a30 = 10 – 87 = -77

Hence, option (C) is correct.

(ii) 11th term of the A.P.: -3, -½, 2, ……, is:

(A) 28                  (B) 22                    (C) -38               (D) -48½

Here a = -3, d = - ½ – (-3) = 5/2 and n = 11,

Therefore, substituting the value in an = a + (n – 1)d,

a11 = -3 + (11 – 1)×5/2

a11 = -3 + 25 = 22

Hence, option (B) is correct.

3. In the following APs, find the missing terms in the boxes:

(i) 2, _, 26

Given, a = 2 and a3 = 26,

To find a2

a3 = a + (3 – 1)d = 26 [given that]

2 + 2d = 26

d = 12

Therefore, a2 = a + (2 – 1)d = 2 + 12 = 14

(ii) _, 13, _, 3

Given, a2 = 13  and a4 = 3,

To find a1 and a3

a2 = a + (2 – 1)d = 13  [given that]

a + d = 13

a = 13 – d  …………(1)

and a4 = 3

a + (4 – 1)d = 3

a + 3d = 3

Substituting the value of a in equation (1),

13 – d + 3d = 3

13 + 2d = 3

d = -5

Substituting the value of d in equation (1),

a = 13 – (-5)

a = 13 + 5

a = 18

Therefore a1 = 18 and a3 = a + (3 – 1)d = 18 + 2(-5) = 8

(iii) 5, _, _, 9 ½

Given, a = 5  and a4 = 9 ½,

To find a2 and a3

a4 = a + (4 – 1)d = 9 ½ [given that]

5 + 3d = 19/2

3d = 19/2 – 5

3d = 9/2

d = 3/2

Therefore a2 = a + d = 5 + 3/2 =  6 ½ and a3 = a + 2d = 5 + 2(3/2) = 8

(iv) -4,_, _, _, _, 6

Given, a = - 4 and a6 = 6,

To find a2, a3, a4 and a5

a6 = a + (6 – 1)d = 6  [given that]

-4 + 5d = 6

5d = 10

d = 2

Therefore a2 = a + d = -4 + 2 = -2

a3 = a + 2d = -4 + 2×2 = 0

a4 = a + 3d = -4 + 3×2 = 2

a5 = a + 4d = -4 + 4×2 = 4

(v) _, 38, _, _, _, 22

Given, a2 = 38 and a6 = -22,

To find a1, a3, a4 and a5

a2 = a + (2 – 1)d = 38  [given that]

a + d = 38

a = 38 – d  …………(1)

and a6 = -22

a + 5d = -22

Substituting the value of a in equation (1),

38  - d + 5d = -22

38 + 4d = -22

d = -15

Substituting the value of d in equation (1),

a = 38 – (-15) = 53

Therefore, a1 = a = 53

a3 = a + 2d = 53 + 2×-15 = 23

a4 = a + 3d = 53 + 2×-15 = 8

a5 = a + 4d = 53 + 2×-15 = -7

4. Which term of the A.P.: 3, 8, 13, 18, … is 78?

Given a = 3 and d = 8 – 3 = 5,

Let, nth term of the A.P. is 78.

Therefore, an = 78

a + (n – 1)d = 78

3 + (n – 1)5 = 78

(n – 1)5 = 75

n – 1 = 15

n  = 16

Hence, 16th term of the A.P.: 3, 8, 13, 18, … is 78

5. Find the number of terms in each of the following APs?

(i) 7, 13, 19, …, 205

Here a = 7 and d = 13 – 7 = 6

Let there be n terms in the A.P.

Therefore, an = 205

a + (n – 1)d = 205

7 + (n – 1)6 = 205

(n – 1)6 = 198

n – 1 = 33

n  = 34

Hence, there are 34 terms in A.P.

(ii) 18, 15 ½ , 13, …… , -47

Here a = 18 and d = 15 ½ - 18 = -5/2

Let there be n terms in the A.P.

Therefore, an = -47

a + (n – 1)d = -47

18 + (n – 1)(-5/2) = -47

(n – 1)(-5/2) = -65

n – 1 = 26

n = 27

Hence, there are 27 terms in A.P.

6. Check whether -150 is a term of the  A.P.: 11, 8, 5, 2, …?

Here a = 11 and d = 8 – 11 = -3

Let, the nth term of the A.P. is -150.

Therefore an = -150

a + (n – 1)d = -150

11 + (n – 1)×-3 = -150

11 – 3n + 3 = -150

-3n = -164

n = 164/3 =

Here n is not an integer. Hence, -150 is not the term of A.P.: 11, 8, 5, 2, ….

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Here, a11 = 38  and a16 = 73,

To find a31

a11 = a + (11 – 1)d = 38  [given that]

a + 10d = 38

a = 38 – 10d          ……………(1)

and a16 = 73

a + 15d = 73

Substituting the value of a in equation (1),

38 – 10d + 15d = 73

5d = 35

d = 7

Substituting the value of d in equation (1),

a = 38 – 10×7 = -32

Therefore a31 = a + 30d = -32 + 30×7 = 178

Hence, the 31st term of A.P is 178.

8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Given, a3 = 12 and a50 = 106,

To find a29

a3 = a + (3 – 1)d = 12  [given that]

a + 2d = 12

a = 12 – 2d   …………(1)

and a50 = 106

a + 49d = 106

Substituting the value of a in equation (1),

12 – 2d + 49d = 106

47d = 94

d = 2

Substituting the value of d in equation (1),

a = 12 – 2×2 = 8

Therefore a29 = a + 28d = 8 + 28×2 = 64

Hence, 29th term of the A.P. is 64.

9. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?

Given, a3 = 4 and a9 = -8,

To find n: where an = 0

a3 = a + (3 – 1)d = 4   [given that]

a + 2d = 4

a = 4 – 2d  …………(1)

and a9 = -8

a + 8d = -8

Substituting the value of a in equation (1),

4 – 2d + 8d = -8

6d  = - 12

d = -2

Substituting the value of d in equation (1),

a = 4 – 2×-2 = 8

So putting the value in an = 0,

an = a + (n – 1)d = 0

8 + (n – 1)×-2 = 0

n – 1 = 4

n = 5

So, A.P. 5th term of is zero.

10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Let the first term = a and common difference = d

According to question,

a17 = a10 + 7

a + 16d = a + 9d + 7

7d = 7

d = a

So this A.P. The common difference of is 1.

11. Which term of the A.P.: 3, 15, 27, 39, … will be 132 more than its 54th term?

First term = 3 and common difference = 15 – 3 = 12

Let the nth term of this A.P be 132 more than its 54th term.

According to question,

an = a54 + 132

a + (n – 1)d = a + 53d + 132

(n – 1)12 = 53×12 + 132

(n – 1)12 = 768

n – 1 = 768/12 = 64

n = 65

Hence, 65th term of the  A.P.: 3, 15, 27, 39, … will be 132 more than its 54th term.

12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Let the first term of the first AP = A and the common difference = d

and the first term of the second AP = a and the common difference = d

Difference between their 100th term = A100 – a100 = 100

(A + 99d) – (a + 99d) = 100

A – a = 100

Difference between their 1000th term = A1000 – a1000

(A + 999d) – (a + 999d)

A – a

⇒ 100        [ A – a = 100]

Hence, these A.P. The difference of 1000th terms will be 100.

13. How many three-digit numbers are divisible by 7?

Three digit numbers divisible by 7: 105, 112, 119, … , 994

Let the total number of three digit numbers divisible by 7 be n.

Given that, a = 105 and d = 112 – 105 = 7,

To find n: where an = 994

an = a + (n – 1)d = 994   [given that]

105 + (n – 1)7 = 994

7(n – 1) = 889

n – 1 = 889/7

n = 128

Hence, 128 three digit numbers are divisible by 7.

14. How many multiples of 4 lie between 10 and 250?

Multiples of 4 between 10 and 250: 12, 16, 20, … , 248

Let there be total n multiples of 4 between 10 and 250.

Given that, a = 12  and d = 16 – 12 = 4,

To find n: where an = 248

an = a + (n – 1)d = 248   [given that]

12 + (n – 1)4 = 248

4(n – 1) = 236

n – 1 = 236/4

n – 1 = 59

n = 60

Hence, there are 60 multiples of 4 between 10 and 250.

15. For what value of n, are the nth terms of two APs: 63, 65, 67, … and 3, 10, 17, …  equal?

Let the first term of the first AP = A = 63 and the common difference = D = 65 – 63 = 2

Therefore An = A + (n – 1)D

An = 63 + (n – 1)2

and the first term of the second AP = a = 3 and the common difference = d = 10 – 3 = 7

Therefore, an = a + (n – 1)d

an = 3 + (n – 1)7

According to question,

An = an

63 + (n – 1)2 = 3 + (n – 1)7

63 + 2n – 2 = 3 + 7n – 7

65 = 5n

n = 13

Hence, the 13th term of both the APs are equal.

16. Determine the AP Whose third term is 16 and the 7th term exceeds the 5th term by 12.

Let the first term of AP = a and the common difference = d

Third term = 16

a3 = 16

a + 2d = 16   ………(1)

7th term exceeds the 5th term by 12,

a7 = a5 + 12

a + 6d = a + 4d + 12

2d = 12

d = 6

Substituting the value of d in equation (1),

a + 2×6 = 16

a = 4

Hence, A.P. = a, a + d, a + 2d, …….. = 4, 10, 16, ……

17. Find the 20th term from the last term of the A.P.: 3, 8, 13, … , 253.

The 20th term from the last term of the AP: 3, 8, 13, … , 253 = the 20th term from the beginning of the AP: 253, … , 13, 8, 3

In A.P.: 253, … , 13, 8, 3, the first term = 253 and the common difference = 3 – 8 = -5

Therefore, a20 = a + 19d

a20 = 253 + 19×-5 = 253 – 95 = 158

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Let the first term of the AP = a and the common difference = d

According to the first condition,

a4 + a8 = 24

a + 3d + a + 7d = 24

2a + 10d = 24

a + 5d = 12

a = 12 – 5d  …………(1)

According to the second condition,

a6 + a10 = 44

a + 5d + a + 9d = 44

2a + 14d = 44

a + 7d = 22

Substituting the value of a in equation (1),

(12 – 5d) + 7d = 22

2d = 10

d = 5

Substituting the value of d in equation (1),

a = 12 – 5×5

a = -13

इस A.P. के प्रथम तीन पद: a, a + d, a + 2d = -13, -8, -3 हैं।

19. Subba Rao started work in 1995 at an annual salary of5000 and received and increment of ₹200 each year. In which year did his income reach ₹7000?

Starting monthly salary = a = ₹5000 And increment every year (common difference) = d = ₹200

Let his salary become ₹7000 in n years.

Therefore, an = 7000

a + (n – 1)d = 7000

5000 + (n – 1)(200) = 7000

(n – 1)(200) = 2000

n – 1 = 10

n = 11

Hence, in the 11th year his salary became ₹7000.

20. Ramkali save ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the nth week , her weekly savings become ₹20.75, find n.

First week savings = a = ₹5 and increased weekly savings (common difference) = d = ₹1.75

The weekly savings in n years become ₹20.75.

Therefore, an = 20.75

a + (n – 1)d = 20.75

5 + (n – 1)(1.75) = 20.75

(n – 1)(1.75) = 15.75

n – 1 = 15.75/1.75 = 9

n = 10

Hence, the value of n is 10.