# Arithmetic Progressions-Exercise 5.2-Class 10

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 ——

Solution:

1. a = 7, d = 3, n = 8, an = ?

We know that, For an A.P.

an = a + (n − 1) d

= 7 + (8 − 1) 3

= 7 + (7) 3

= 7 + 21 = 28

Hence, an = 28

1. Given that a = −18, n = 10, an = 0, d = ?

We know that, an = a + (n − 1) d

0 = − 18 + (10 − 1) d

18 = 9d

d = 2

III. Given that d = −3, n = 18, an = −5

We know that, an = a + (n − 1) d

−5 = a + (18 − 1) (−3)

−5 = a + (17) (−3)

−5 = a − 51

a = 51 − 5 = 46

Hence, a = 46

IV. a = −18.9, d = 2.5, an = 3.6, n = ?

We know that, an = a + (n − 1) d

3.6 = − 18.9 + (n − 1) 2.5

3.6 + 18.9 = (n − 1) 2.5

22.5 = (n − 1) 2.5

(n-1) = (22.5)/(2.5)

n-1 = 10

n = 9

V. a = 3.5, d = 0, n = 105, an = ?

We know that, an = a + (n − 1) d

an = 3.5 + (105 − 1) 0

an = 3.5 + 104 × 0

an = 3.5

Hence, an = 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

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

(A). 28 (B). 22 (C). − 38 (D).-48½

Solution:

(i) Given that A.P. 10, 7, 4, …

First term, a = 10

Common difference, d = a2 − a1 = 7 − 10 = −3

We know that, an = a + (n − 1) d

a30 = 10 + (30 − 1) (−3)

a30 = 10 + (29) (−3)

a30 = 10 − 87 = −77

Hence, the correct answer is C

(ii)Given that, A.P. -3,-½,2

First term a = −3

Common difference, d = a2 – a­1 = -½ -(-3) = -½+3=5/2

We know that,

an = a + (n − 1) d

a11 = -3 + (11 – 1)(5/2)

a11 = -3 + (50/2)

a11 = -3 + 25 = +22

3. In the following APs find the missing term in the boxes

(i) 2, □, 26

(ii) □, 13, □, 3

(iii) 5, □, □, 9½

(iv) -4, □, □, □, □, 6

(v) □, 38, □, □, □, -22

Solution:

(i) 2, □, 26

For this A.P., a = 2 and a3 = 26

We know that, an = a + (n − 1) d

a3 = 2 + (3 − 1) d

26 = 2 + 2d

24 = 2d

d = 12

a2 = 2 + (2 − 1) 12 = 14

Therefore, 14 is the missing term.

(ii) For this A.P., a2 = 13 and a4 = 3

We know that, an = a + (n − 1) d

a2 = a + (2 − 1) d

13 = a + d …………………………(I)

a4 = a + (4 − 1) d

3 = a + 3d ………………………..(II)

On subtracting (I) from (II), we obtain

−10 = 2d

d = −5

From equation (I), we obtain 13 = a + (−5)

a = 18

a3 = 18 + (3 − 1) (−5)

= 18 + 2 (−5)

= 18 − 10 = 8 Therefore, the missing terms are 18 and 8 respectively.

(iii) For this A.P.,

a = 5

a4 = 9(1/2) = 19/2 Therefore, the missing terms are 13/2 and 8 respectively.

(iv) For this A.P., a = −4 and a6 = 6

We know that, an = a + (n − 1) d

a6 = a + (6 − 1) d

6 = − 4 + 5d

10 = 5d

d = 2

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

Therefore, the missing terms are −2, 0, 2, and 4 respectively.

(v) For this A.P., a2 = 38 and a6 = −22

We know that an = a + (n − 1) d

a2 = a + (2 − 1) d

38 = a + d ………………………(1)

a6 = a + (6 − 1) d

−22 = a + 5d …………………….(2)

On subtracting equation (1) from (2), we obtain,

− 22 − 38 = 4d

−60 = 4d

d = −15

a = a2 − d = 38 − (−15) = 53

a3 = a + 2d = 53 + 2 (−15) = 23

a4 = a + 3d = 53 + 3 (−15) = 8

a5 = a + 4d = 53 + 4 (−15) = −7

Therefore, the missing terms are 53, 23, 8, and −7 respectively.

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

Solution:

3, 8, 13, 18, …

For this A.P., a = 3 and d = a2 – a1 = 8 − 3 = 5

Let nth term of this A.P. be 78.

an = a + (n − 1) d

78 = 3 + (n − 1) 5

75 = (n − 1) 5

(n − 1) = 15

n = 16

Hence, 16th term of this A.P. is 78.

5. Find the number of terms in each of the following A.P.

(i) 7, 13, 19, …, 205

(ii) 18, 15 1/2 , 13, … , −47

Solution:

(i). 7, 13, 19, …, 205

For this A.P., a = 7 and d = a2 – a1 = 13 − 7 = 6

Let there are n terms in this A.P. an = 205

We know that an = a + (n − 1) d

Therefore, 205 = 7 + (n − 1) 6

198 = (n − 1) 6

33 = (n − 1)

n = 34

Therefore, this given series has 34 terms in it.

(ii). 18, 15 1/2 , 13, … , −47

For this A.P.,

a = 18 and d = a2 – a1 = 15(1/2) − 18 = (31-36)/2 = -5/2

Let there are n terms in this A.P. Therefore, an = −47 and we know that, Therefore, this given A.P. has 27 terms in it.

1. Check whether − 150 is a term of the A.P. 11, 8, 5, 2, …

Solution:

For this A.P., a = 11 and d = a2 – a1 = 8 − 11 = −3

Let −150 be the nth term of this A.P. We know that,

an = a + (n – 1)d

-150 = 11 + (n – 1)( -3)

-150 = 11 – 3n + 3

-164 = -3n

n = 164/3

Clearly, n is not an integer. Therefore, −150 is not a term of this A.P.

1. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.

Solution: Given that, a11 = 38 and a16 = 73

We know that, an = a + (n − 1) d

a11 = a + (11 − 1) d

38 = a + 10d ………………………..(1)

Similarly, a16 = a + (16 − 1) d

73 = a + 15d ………………………….(2)

On subtracting (1) from (2), we obtain

35 = 5d

d = 7

From equation (1), 38 = a + 10 × (7)

38 − 70 = a

a = −32

a31 = a + (31 − 1) d

= − 32 + 30 (7)

= − 32 + 210

= 178

Hence, 31st term is 178.

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

Solution:

Given that, a3 = 12 and a50 = 106

We know that, an = a + (n − 1) d

a3= a + (3 − 1) d

12 = a + 2d …………………………..(I)

Similarly, a50 = a + (50 − 1) d

106 = a + 49d ………………………(II)

On subtracting (I) from (II), we obtain

94 = 47d

d = 2

From equation (I), we obtain

12 = a + 2 (2)

a = 12 − 4 = 8

a29 = a + (29 − 1) d

a29 = 8 + (28)2

a29 = 8 + 56 = 64

Therefore, 29th term is 64.

9: If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero?

Solution:

Given that, a3 = 4 and a9 = −8

We know that, an = a + (n − 1) d

a3 = a + (3 − 1) d

4 = a + 2d ……………………(I)

a9 = a + (9 − 1) d

−8 = a + 8d …………………(II)

On subtracting equation (I) from (II), we obtain

−12 = 6d

d = −2

From equation (I), we obtain

4 = a + 2 (−2)

4 = a − 4

a = 8

Let n th term of this A.P. be zero.

an = a + (n − 1) d

0 = 8 + (n − 1) (−2)

0 = 8 − 2n + 2

2n = 10

n = 5

Hence, 5th term of this A.P. is 0.

10. If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.

Solution:

We know that,

For an A.P., an = a + (n − 1) d

a17 = a + (17 − 1) d

a17  = a + 16d

Similarly, a10 = a + 9d

It is given that a17 – a10 = 7 (a + 16d) − (a + 9d) = 7 7d = 7 d = 1 Therefore, the common difference is 1.

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

Solution:

Given A.P. is 3, 15, 27, 39, …

a = 3

d = a2 – a1 = 15 − 3 = 12

a54 = a + (54 − 1) d

= 3 + (53) (12)

= 3 + 636

= 639

Now a54 +132 = 639 + 132 = 771

We have to find the term of this A.P. which is 771.

Let n th term be 771.

an = a + (n − 1) d

771 = 3 + (n − 1) 12

768 = (n − 1) 12

(n − 1) = 64

n = 65

Therefore, 65th term was 132 more than 54th term.

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

Solution:

Let the first term of these A.P.s be a1 and a2 respectively and the common difference of these A.P.s be d. For first A.P.,

a100 = a1 + (100 − 1) d

= a1 + 99d

a1000 = a1 + (1000 − 1) d

a1000 = a1 + 999d

For second A.P., a100 = a2 + (100 − 1) d

= a2 + 99d

a1000 = a2 + (1000 − 1) d

= a2 + 999d

Given that, difference between 100th term of these A.P.s = 100 Therefore, (a1 + 99d) − (a1 + 99d) = 100

a1 – a1 = 100 ……………………(1)

Difference between 1000th terms of these A.P.s

(a1 + 999d) − (a1 + 999d) = a1 – a2

From equation (1), This difference, a1 – a2 = 100

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

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

Solution:

First three-digit number that is divisible by 7 = 105

Next number = 105 + 7 = 112

Therefore, 105, 112, 119, …

All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.

The maximum possible three-digit number is 999.

When we divide it by 7, the remainder will be 5.

Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7. The series is as follows. 105, 112, 119, …, 994 Let 994 be the nth term of this A.P.

a = 105, d = 7 and an = 994. n = ?

an = a + (n − 1) d

994 = 105 + (n − 1) 7

889 = (n − 1) 7

(n − 1) = 127

n = 128

Therefore, 128 three-digit numbers are divisible by 7

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

Solution:

First multiple of 4 that is greater than 10 is 12.

Next will be 16. Therefore, 12, 16, 20, 24, …

All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.

When we divide 250 by 4, the remainder will be 2.

Therefore, 250 − 2 = 248 is divisible by 4.

The series is as follows. 12, 16, 20, 24, …, 248 Let 248 be the n th term of this A.P.

a = 12

d = 14

an = 248

an = a + (n – 1)d

248 = 12 + (n-1)4

236/4 = n – 1

59 = n – 1

n = 60

Therefore, there are 60 multiples of 4 between 10 and 250

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

Solution:

For AP: 63, 65, 67, …

a = 63 and d = a2 − a1 = 65 − 63 = 2

n th term of this A.P.

an = a + (n − 1) d

an = 63 + (n − 1) 2

= 63 + 2n − 2

an = 61 + 2n ……………(1)

For AP:3, 10, 17, … a = 3 and d = a2 – a1 = 10 − 3 = 7

n th term of this A.P. = 3 + (n − 1) 7

an = 3 + 7n − 7

an = 7n − 4 …………………………………(2)

It is given that, n th term of these A.P.s are equal to each other. Equating both these equations, we obtain

61 + 2n = 7n − 4

61 + 4 = 5n

5n = 65

n = 13

Therefore, 13th terms of both these A.P.s are equal to each other.

16. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5 th term by 12

Solution:

a3 = 16 a + (3 – 1) d

= 16a + 2d = 16 ………………………….(1)

a7 – a5 = 12

[a+ (7 − 1) d] − [a + (5 − 1) d]= 12

(a + 6d) − (a + 4d) = 12

2d = 12

d = 6

From equation (1), we obtain

a + 2 (6) = 16

a + 12 = 16

a = 4

Therefore, A.P. will be 4, 10, 16, 22, …

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

Solution:

Given A.P. is 3, 8, 13, …, 253

Common difference for this A.P. is 5.

Therefore, this A.P. can be written in reverse order as 253, 248, 243, …, 13, 8, 5

For this A.P., a = 253 d = 248 − 253 = −5

n = 20

a20 = a + (20 − 1) d

a20 = 253 + (19) (−5)

a20 = 253 − 95

a = 158

Therefore, 20 th term from the last term is 158.

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

Solution:

We know that, an = a + (n − 1) d

a4 = a + (4 − 1) d

a4 = a + 3d

Similarly, a8 = a + 7d

a6 = a + 5d

a10 = a + 9d

Given that, a4 + a8 = 24

a + 3d + a + 7d = 24

2a + 10d = 24

a + 5d = 12 …………………………….(1)

a6 + a10 = 44 a + 5d + a + 9d = 44 2a + 14d = 44 a + 7d = 22 …………………………….(2)

On subtracting equation (1) from (2), we obtain

2d = 22 − 12

2d = 10

d = 5

From equation (1), we obtain a + 5d = 12

a + 5 (5) = 12

a + 25 = 12

a = −13

a2 = a + d = − 13 + 5 = −8

a3 = a2 + d = − 8 + 5 = −3

Therefore, the first three terms of this A.P. are −13, −8, and −3.

19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Solution:

It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.

Therefore, the salaries of each year after 1995 are  5000, 5200, 5400, …

Here, a = 5000 and d = 200

Let after n th year, his salary be Rs 7000.

Therefore, an = a + (n − 1) d

7000 = 5000 + (n − 1) 200

200(n − 1) = 2000

(n − 1) = 10

n = 11

Therefore, in 11th year, his salary will be Rs 7000.

20. Ramkali saved 𝑅𝑠 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the n th week, her week, her weekly savings become Rs 20.75, find n.

Solution:

Given that, a = 5, d = 1.75 and an = 20.75. n = ?

an = a + (n − 1) d

20.75 = 5 + (n – 1)1.75

15.75 = (n-1)1.75

(n-1) = (15.75)/(1.75) = 1575/175

63/7 = 9

n – 1 = 9

n = 10

Hnce, n = 10