# Arithmetic Progressions Exercise 5.2

Arithmetic Progressions Exercise 5.2

Page No: 105

1. Fill in the blanks in the following table, given that a is the first term, dthe common difference and an the nthterm 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) 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

(ii) Given that
a= −18, n = 10, an = 0, d= ?
We know that,
ana + (n − 1) d
0 = − 18 + (10 − 1) d
18 = 9d
d = 18/9 = 2
Hence, common difference, = 2

(iii) Given that
d= −3, n = 18, an = −5
We know that,
ana + (n − 1) d
−5 = a + (18 − 1) (−3)
−5 = a + (17) (−3)
−5 = − 51
a= 51 − 5 = 46
Hence,a = 46

(iv) a = −18.9, d = 2.5, an = 3.6, n= ?
We know that,
ana + (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 = 9
n = 10
Hence, n = 10

(v) a = 3.5, d = 0, n = 105, an = ?
We know that,
ana + (n − 1) d
an= 3.5 + (105 − 1) 0
an= 3.5 + 104 × 0
an= 3.5
Hence,an = 3.5

Page No: 106

Choose the correct choice in the following and justify
(i) 30thterm of the A.P: 10, 7, 4, …, is
(A) 97 (B) 77 (C) −77 (D.) −87

(ii) 11th term of the A.P. -3, -1/2, ,2 .... is
(A) 28 (B) 22 (C) - 38 (D) (i) Given that
A.P. 10, 7, 4, …
First term, a = 10
Common difference, d = a2 − a= 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 option C.

(ii) Given that A.P. is -3, -1/2, ,2 ...
First term a = - 3
Common difference, d = a2 − a1 = (-1/2) - (-3)
= (-1/2) + 3 = 5/2
We know that, an = a + (n − 1) d
a11 = 3 + (11 -1)(5/2)
a11 = 3 + (10)(5/2)
a11 = -3 + 25
a11 = 22
Hence, the answer is option B.

3. In the following APs find the missing term in the boxes. (i) For this A.P.,

a = 2
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 get
- 10 = 2d
d = - 5
From equation (i), we get
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.,
= 5 and
a4 = 19/2
We know that, an = a + (n − 1) d
a4 = a + (4 - 1) d
19/2 = 5 + 3d
19/2 - 5 = 3d3d = 9/2
d = 3/2

a2 = a + (2 - 1) d
a2 = + 3/2
a2 = 13/2

a3 = a + (3 - 1) d

a3 = 5 + 2×3/2

a3 = 8

Therefore, the missing terms are 13/2 and 8 respectively.

(iv) For this A.P.,
a= −4 and
a6= 6
We know that,
ana + (n − 1) d
a6= a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d= 2
a2a + d = − 4 + 2 = −2
a3a + 2d = − 4 + 2 (2) = 0
a4a + 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
a6= −22
We know that
ana + (n − 1) d
a2a + (2 − 1) d
38 = a + d ... (i)
a6a + (6 − 1) d
−22 = a + 5d ... (ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d
d= −15
aa2 − d = 38 − (−15) = 53
a3+ 2= 53 + 2 (−15) = 23
a4a + 3d = 53 + 3 (−15) = 8
a5a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.

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

3, 8, 13, 18, …
For this A.P.,
a = 3
d = a2− a1 = 8 − 3 = 5
Let nthterm of this A.P. be 78.
ana + (n − 1) d
78 = 3 + (n − 1) 5
75 = (n − 1) 5
(n − 1) = 15
n = 16
Hence, 16thterm 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, , 13,...., -47

(i) For this A.P.,
a= 7
da2 − a1 = 13 − 7 = 6
Let there are n terms in this A.P.
an= 205
We know that
ana + (n − 1)
Therefore, 205 = 7 + (− 1) 6
198 = (n − 1) 6
33 = (n − 1)
n= 34
Therefore, this given series has 34 terms in it.

(ii) For this A.P.,
a = 18 Let there are n terms in this A.P.
an = 205

an = a + (n − 1) d

-47 = 18 + (n - 1) (-5/2)

-47 - 18 = (n - 1) (-5/2)
-65 = (n - 1)(-5/2)
(n - 1) = -130/-5
(n - 1) = 26
= 27
Therefore, this given A.P. has 27 terms in it.

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

For this A.P.,
a = 11
d = a2− a1 = 8 − 11 = −3
Let −150 be thenth 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.

7. Find the 31stterm of an A.P. whose 11th term is 38 and the 16thterm is 73.

Given that,
a11 = 38
a16 = 73
We know that,
ana + (n − 1) d
a11 =+ (11 − 1) d
38 = a + 10d ... (i)
Similarly,
a16 =a + (16 − 1) d
73 = a + 15d ... (ii)
On subtracting (i) from(ii), we get
35 = 5d
d = 7
From equation (i),
38 = a + 10 × (7)
38 − 70 = a
a = −32
a31 =+ (31 − 1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, 31stterm 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.

Given that,
a3 = 12
a50 = 106
We know that,
ana + (n − 1) d
a3 =a + (3 − 1) d
12 = a + 2d ... (i)
Similarly, a50a + (50 − 1) d
106 = a + 49d ... (ii)
On subtracting (i) from(ii), we get
94 = 47d
d = 2
From equation (i), we get
12 = a + 2 (2)
a = 12 − 4 = 8
a29 =a + (29 − 1) d
a29 = 8 + (28)2
a29 = 8 + 56 = 64
Therefore, 29thterm is 64.

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

Given that,

a3 = 4
a9 = −8
We know that,
ana + (n − 1) d
a3 = + (3 − 1) d
4 = a + 2d ... (i)
a9 =+ (9 − 1) d
−8 = a + 8d ... (ii)
On subtracting equation(i) from (ii), we get,
−12 = 6d
d = −2
From equation (i), we get,
4 = + 2 (−2)
4 = a − 4
a = 8
Let nthterm of this A.P. be zero.
a=+ (− 1) d
0 = 8 + (n − 1) (−2)
0 = 8 − 2n+ 2
2= 10
n = 5
Hence, 5thterm of this A.P. is 0.

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

We know that,
For an A.P., ana + (n − 1) d
a17 =a + (17 − 1) d
a17 =a + 16d
Similarly, a10a + 9d
It is given that
a17 −a10 = 7
(a + 16d) − (a + 9d) = 7
7d = 7
d = 1
Therefore, the common difference is 1.

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

Given A.P. is 3, 15, 27, 39, …
= 3
d = a2− a1 = 15 − 3 = 12
a54 =a + (54 − 1) d
= 3 + (53) (12)
= 3 + 636 = 639
132 + 639 = 771
We have to find the term of this A.P. which is 771.
Let nthterm be 771.
ana + (n − 1) d
771 = 3 + (n − 1) 12
768 = (n − 1) 12
(n − 1) = 64
n = 65
Therefore, 65thterm was 132 more than 54th term.

Or

Let nthterm be 132 more than 54th term.
n = 54 + 132/2
= 54 + 11 = 65th term

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

Let the first term of these A.P.s be a1 and a2respectively and the common difference of these A.P.s be d.
For first A.P.,
a100a1 + (100 − 1) d
=a1 + 99d
a1000a1 + (1000 − 1) d
a1000a1 + 999d
For second A.P.,
a100a2 + (100 − 1) d
=a2 + 99d
a1000a2 + (1000 − 1) d
=a2 + 999d
Given that, difference between
100th term of these A.P.s = 100
Therefore, (a1+ 99d) − (a2 + 99d) = 100
a1 −a2 = 100 ... (i)
Difference between 1000th terms of these A.P.s
(a1 + 999d) − (a2 + 999d) = a1− a2
From equation (i),
This difference, a1− a= 100
Hence, the difference between 1000th terms of these A.P. will be 100.

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

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
an= 994
n = ?
ana + (n − 1) d
994 = 105 + (n − 1) 7
889 = (n − 1) 7
(− 1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.

Or

Three digit numbers which are divisible by 7 are 105, 112, 119, .... 994 .
These numbers form an AP with a = 105 and d = 7.
Let number of three-digit numbers divisible by 7 be nan = 994
⇒ a + (n - 1) d = 994
⇒ 105 + (n - 1) × 7 = 994
⇒7(n - 1) = 889
⇒ n - 1 = 127
⇒ n = 128

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

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 nthterm of this A.P.
a = 12
d = 4
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.

Or

Multiples of 4 lies between 10 and 250 are 12, 16, 20, ...., 248.
These numbers form an AP with a = 12 and d = 4.
Let number of three-digit numbers divisible by 4 be nan = 248
⇒ a + (n - 1) d = 248
⇒ 12 + (n - 1) × 4 = 248
⇒4(n - 1) = 248
⇒ n - 1 = 59
⇒ n = 60

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

63, 65, 67, …
a = 63
d = a2 − a1 = 65 − 63 = 2
nthterm of this A.P. =  an = a + (n− 1) d
an= 63 + (n − 1) 2 = 63 + 2n − 2
an = 61 + 2n ... (i)
3, 10, 17, …
a = 3
d = a2− a1 = 10 − 3 = 7
nthterm of this A.P. = 3 + (n − 1) 7
an= 3 + 7n − 7
an= 7n − 4 ... (ii)
It is given that, nthterm 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, 13thterms 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 5th term by 12.

a3 = 16

a + (3 − 1) d = 16
a + 2d = 16 ... (i)
a7 − a5 = 12
[a+ (7 − 1) d] − [+ (5 − 1) d]= 12
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
From equation (i), we get,
a + 2 (6) = 16
a + 12 = 16
a = 4
Therefore, A.P. will be
4, 10, 16, 22, …

Page No: 107

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

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
= 20
a20 =a + (20 − 1) d
a20 = 253 + (19) (−5)
a20 = 253 − 95
a = 158
Therefore, 20thterm 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.

We know that,
an = a + (− 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 + + 7d = 24
2a + 10d = 24
a + 5d = 12 ... (i)
a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 ... (ii)
On subtracting equation (i) from (ii), we get,
2d = 22 − 12
2d = 10
d = 5
From equation (i), we get
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?

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
d = 200
Let after nthyear, his salary be Rs 7000.
Therefore, ana + (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 nth week, her week, her weekly savings become Rs 20.75, find n.

Given that,
a = 5
= 1.75
a= 20.75
n = ?
ana + (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
Hence, n is 10.