NCERT solutions for class 10 Maths chapter Arithmetic Progressions (Exercise 5.2) Ex 5.2 Question 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.

i738...
ii-18...100
iii...-318-5
iv-18.92.5...3.6
v3.50105...

Solutions:

(i)a = 7

d = 3

n= 8,

Putiing the values in

an = a+(n−1)d

an = 7+(8 −1) 3 =  7+(7) 3 = 7+21 = 28

an = 28

(ii) a = -18

d = ?

n = 10

an = 0

Putiing the values in,

an = a+(n−1)d

0 = − 18 +(10−1)d

18 = 9d

d = 18/9 = 2

(iii) a =?

d = -3

n = 18

an = -5

Putiing the values in,

an = a+(n−1)d

−5 = a+(18−1) (−3)

−5 = a+(17) (−3)

−5 = a−51

a = 51−5 = 46

(iv) a = -18.9

d = 2.5

n =?

an = 3.6

Putiing the values in,

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 = 9

n = 10

(v)  a = 3.5

d = 0

n = 105

an = ?

Putiing the values in,

an = a+(n −1)d

an = 3.5+(105−1) 0

an = 3.5+104×0

an = 3.5

Ex 5.2 Question 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, -1/2, 2 …. is
(A) 28 (B) 22 (C) – 38 (D) -48½

Solutions:

(i) A.P. = 10, 7, 4, …

a = 10

d = a2 − a= 7−10 = −3

Putting the values in

an = a +(n−1)d

a30 = 10+(30−1)(−3)

a30 = 10+(29)(−3)

a30 = 10−87 = −77

Hence, option C is correct.

(ii)A.P. = -3, -1/2, ,2 …

a = – 3

d = a2 − a1 = (-1/2) -(-3)

⇒(-1/2) + 3 = 5/2

Putting the values in

an = a+(n−1)d

a11 = 3+(11-1)(5/2)

a11 = 3+(10)(5/2)

a11 = -3+25

a11 = 22

Hence, option B is correct.

Ex 5.2 Question 3.

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

2....26
....13.....3
5........
-4..................6
....38.............-22

Solution:

(i)a = 2

a3 = 26

Putting the values in

an = a+(n −1)d

a3 = 2+(3-1)d

26 = 2+2d

24 = 2d

d = 12

a2 = 2+(2-1)12

= 14

(ii)a2 = 13

a4 = 3

Putting the values in

an = a+(n−1) d

a2 = a +(2-1)d

13 = a+d ………………. (i)

a4 = a+(4-1)d

3 = a+3d ………….. (ii)

Subtracting equation (i) from (ii),

– 10 = 2d

d = – 5

From equation (i), putting the value of d,

13 = a+(-5)

a = 18

a3 = 18+(3-1)(-5)

= 18+2(-5) = 18-10 = 8

(iii)= 5 and

a4 = 19/2

Putting values in

an = a+(n−1)d

a4 = a+(4-1)d

19/2 = 5+3d

(19/2) – 5 = 3d

3d = 9/2

d = 3/2

a2 = a+(2-1)d

a2 = 5+3/2

a2 = 13/2

a3 = a+(3-1)d

a3 = 5+2×3/2

a3 = 8

(iv) a = −4

a6 = 6

Putting the values in,

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

(v)a2 = 38

a6 = −22

Putting values in

an = a+(n −1)d

a2 = a+(2−1)d

38 = a+d ……………………. (i)

a6 = a+(6−1)d

−22 = a+5d ………………….(ii)

Subtracting equation (i) from (ii),

− 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

Ex 5.2 Question 4.

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

Solutions:

Given the A.P. series as3, 8, 13, 18, …

First term, a = 3

Common difference, d = a2 − a1 = 8 − 3 = 5

Let the nth term of given A.P. be 78. Now as we know,

an = a+(n−1)d

Therefore,

78 = 3+(n −1)5

75 = (n−1)5

(n−1) = 15

n = 16

Hence, 16th term of this A.P. is 78.

Ex 5.2 Question 5.

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

(i) 7, 13, 19, …, 205

(ii) 18, 15½,13….., -47

Solution:

(i)  a = 7

d = a2 − a1 = 13 − 7 = 6

an = 205

an = a + (n − 1) d

205 = 7 + (n − 1) 6

198 = (n − 1) 6

33 = (n − 1)

n = 34

(ii) a = 18

d = a2-a= 15½ – 18

d = (31-36)/2 = -5/2

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

n = 27

Ex 5.2 Question 6.

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

Solution:

a = 11

d = a2−a1 = 8−11 = −3

an = -150

an = a+(n−1)d

-150 = 11+(n -1)(-3)

-150 = 11-3n +3

-164 = -3n

n = 164/3

So, n is a fraction not an integer.

Hence, -150 is nota a term of A.P.

Ex 5.2 Question 7.

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

Solution:

11th term, a11 = 38

16th term, a16 = 73

an = a+(n−1)d

a11 = a+(11−1)d

38 = a+10d ………………. (i)

a16 = a +(16−1)d

73 = a+15d …………………… (ii)

Subtracting equation(i) from(ii),

35 = 5d

d = 7

From equation (i),

38 = a+10×(7)

38 − 70 = a

a = −32

a31 = a +(31−1) d

= − 32 + 30 (7)

= − 32 + 210

= 178

31st term =178.

Ex 5.2 Question 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:

3rd term, a3 = 12

50th term, a50 = 106

an = a+(n−1)d

a3 = a+(3−1)d

12 = a+2d …………. (i)

a50 = a+(50−1)d

106 = a+49d ……………. (ii)

Subtracting equation (i) from (ii),

94 = 47d

d = 2

From equation (i),

12 = a+2(2)

a = 12−4 = 8

a29 = a+(29−1) d

a29 = 8+(28)2

a29 = 8+56 = 64

29th term = 64.

Ex 5.2 Question 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:

3rd term, a3 = 4

9th term, a9 = −8

an = a+(n−1)d

a3 = a+(3−1)d

4 = a+2d ……………… (i)

a9 = a+(9−1)d

−8 = a+8d …………… (ii)

Subtracting equation (i) from (ii)

−12 = 6d

d = −2

From equation (i),

4 = a+2(−2)

4 = a−4

a = 8

Let nth term =0

a= a+(n−1)d

0 = 8+(n−1)(−2)

0 = 8−2n+2

2n = 10

n = 5

5th term = 0.

Ex 5.2 Question 10.

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

Solution:

an = a+(n−1)d

a17 = a+(17−1)d

a17 = a +16d

a10 = a+9d

a17 − a10 = 7

(a +16d)−(a+9d) = 7

7d = 7

d = 1

Ex 5.2 Question 11.

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

Solution:

a = 3

d = a2 − a1 = 15 − 3 = 12

an = a+(n−1)d

a54 = a+(54−1)d

⇒3+(53)(12)

⇒3+636 = 639

a54 = 639

Let nth = 771.

an = a+(n−1)d

771 = 3+(n −1)12

768 = (n−1)12

(n −1) = 64

n = 65

65th term = 132.

Ex 5.2 Question 12.

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 = a1 and

Second term = a2

Common difference = d.

For first term-

an = a+(n−1)d

a100 = a1+(100−1)d

= a1 + 99d

a1000 = a1+(1000−1)d

a1000 = a1+999d

For second term –

an = a+(n−1)d

a100 = a2+(100−1)d

= a2+99d

a1000 = a2+(1000−1)d

= a2+999d

The difference between 100th term of the two APs = 100

∴(a1+99d) − (a2+99d) = 100

a1−a2 = 100……….. (i)

Difference between 1000th terms of the two APs = ?

(a1+999d) − (a2+999d) = a1−a2

From equation (i),

a1−a= 100

∴The difference between the1000th terms of the two A.P. = 100.

Ex 5.2 Question 13.

How many three-digit numbers are divisible by 7?

Solution:

First,three-digit number that is divisible by 7

First number = 105

Second number = 105+7 = 112

Third number = 112+7 =119

∴105, 112, 119, …

Let the total number of these numbers be n.

a = 105

d = 112 – 105 = 7

n = ?

an = 994

an = a+(n−1)d

994 = 105+(n−1)7

889 = (n−1)7

(n−1) = 127

n = 128

∴ 128 three-digit numbers are divisible by 7.

Ex 5.2 Question 14.

How many multiples of 4 lie between 10 and 250?

Solution:

Multiples of 4 lie between 10 and 250 = 12, 16, 20, 24, …248.

Let the total number of multiplies of 4  lie between 10 and 250 be n.

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

∴ Total number of multiplies of 4 lie between 10 and 250 = 60

Ex 5.2 Question 15.

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

Solution:

63, 65, 67, …

a = 63

d = a2−a1 = 65−63 = 2

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

an = a+(n−1)d

an = 3+7n−7

an = 7n−4 ………….. (ii)

nth term of these A.P.s are equal.

Equating  equation (i) and (ii)

61+2n = 7n−4

61+4 = 5n

5n = 65

n = 13

∴13th terms of both these A.P.s are equal.

Ex 5.2 Question 16.

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

Solutions:

a3 = 16

a +(3−1)d = 16

a+2d = 16 …………. (i)

7th term exceeds the 5th term by 12.

a7 − a5 = 12

[a+(7−1)d]−[+(5−1)d]= 12

(a+6d)−(a+4d) = 12

2d = 12

d = 6

From equation (i),

a+2(6) = 16

a+12 = 16

a = 4

∴ A.P. =4, 10, 16, 22, …

Ex 5.2 Question 17.

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

Solution:

A.P.  =3,8, 13, …, 253

d= 5.

AP in reverse order –

253, 248, 243, …, 13, 8, 5

For  new AP,

a = 253

d = 248 − 253 = −5

n = 20

an = a+(n−1)d

a20 = a+(20−1)d

a20 = 253+(19)(−5)

a20 = 253−95

a = 158

∴ 20th term from the last term of the AP 3, 8, 13, …, 253.is 158.

Ex 5.2 Question 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:

an = a+(n−1)d

a4 = a+(4−1)d

a4 = a+3d

a8 = a+7d

a6 = a+5d

a10 = a+9d

a4+a8 = 24

a+3d+a+7d = 24

2a+10d = 24

a+5d = 12 …………………… (i)

a6+a10 = 44

a +5d+a+9d = 44

2a+14d = 44

a+7d = 22 ………………………(ii)

Subtracting equation (i) from (ii)

2d = 22 − 12

2d = 10

d = 5

From equation (i)

a+5d = 12

a+5(5) = 12

a+25 = 12

a = −13

a2 = a+d = − 13+5 = −8

a3 = a2+d = − 8+5 = −3

First three terms of this A.P. = −13, −8, and −3.

Ex 5.2 Question 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:

a = 5000

d = 200

an = 7000

an = a+(n−1) d

7000 = 5000+(n−1)200

200(n−1)= 2000

(n−1) = 10

n = 11

∴ 11th year, his salary =Rs 7000.

Ex 5.2 Question 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 weekly savings become Rs 20.75, find n.

Solution:

a = 5

d = 1.75

a= 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