NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

NCERT Solutions Class 10 Maths
Chapter – 5 (Arithmetic Progressions)

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 5 Arithmetic Progressions Exercise 5.2 has been provided here to help the students in solving the questions from this exercise.

Chapter : 5 Arithmetic Progressions

Exercise – 5.2

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the n^th term of the A.P.

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

Solutions –

(i) First term, a = 7
Common difference, d = 3
Number of terms, n = 8,
We have to find the n^th term, a_n = ?
As we know, for an A.P.,
a_n = a+(n−1)d
Putting the values,
⇒ 7 + (8 −1) 3
⇒ 7 + (7) 3
⇒ 7 + 21
⇒ 28
Hence, a_n = 28

(ii) First term, a = -18
Common difference, d = ?
Number of terms, n = 10
n^th term, a_n = 0
As we know, for an A.P.,
a_n = a+(n−1)d
⇒ 0 = − 18 +(10−1)d
⇒ 18 = 9d
⇒ d = 18/9
⇒ d = 2
Hence, common difference, d = 2

(iii) First term, a = ?
Common difference, d = -3
Number of terms, n = 18
n^th term, a_n = -5
As we know, for an A.P.,
a_n = a + (n−1)d
−5 = a+(18−1) (−3)
−5 = a+(17) (−3)
−5 = a−51
a = 51−5
a = 46
Hence, a = 46

(iv) First term, a = -18.9
Common difference, d = 2.5
Number of terms, n = ?
n^th term, a_n = 3.6
As we know, for an A.P.,
a_n = 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
Hence, n = 10

(v) First term, a = 3.5
Common difference, d = 0
Number of terms, n = 105
n^th term, a_n = ?
As we know, for an A.P.,
a_n = a + (n −1)d
a_n = 3.5 + (105 − 1)0
a_n = 3.5 + 104 × 0
a_n = 3.5
Hence, a_n = 3.5

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

Solutions – Given
A.P. = 10, 7, 4, …
First term, a = 10
Common difference, d = a₂ − a₁= 7 − 10 = −3
a_n = a + (n − 1)d
a₃₀ = 10 + (30 − 1)(−3)
a₃₀ = 10 + (29)(−3)
a₃₀ = 10 − 87
a₃₀ = −77
Hence, the correct answer is option C.

(ii) 11^thterm of the A.P. -3, -1/2, ,2 …. is
(A) 28
(B) 22
(C) – 38
(D) $-48\frac{1}{2}$

Solutions – Given
A.P. = -3, -1/2 ,2 …
First term a = – 3
Common difference, d = a₂ − a₁
= (-1/2) – (-3)
= (-1/2) + 3
= 5/2
a_n = 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) 2, $\square$ , 26
(ii) $\square$ , 13, $\square$ , 3
(iii) 5, $\square$ , $\square$ , $9\frac{1}{2}$
(iv) -4, $\square$ , $\square$ , $\square$ , $\square$ , 6
(v) $\square$ , 38, $\square$ , $\square$ , $\square$ , -22

Solutions –
(i) 2, $\square$ , 26
Given
a = 2
a₃ = 26
As we know, for an A.P.,
a_n = a + (n − 1)d
a₃ = 2 + (3 – 1)d
26 = 2 + 2d
24 = 2d
d = 12
a₂ = 2 + (2 – 1)12
a₂ = 14
Therefore, The Series is – 2, 14, 26

(ii) $\square$ , 13, $\square$ , 3
a₂ = 13
a₄ = 3
As we know, for an A.P.,
a_n = a + (n − 1) d
a₂ = a + (2 – 1)d
13 = a + d —————- (i)
a₄ = a + (4 – 1)d
3 = a + 3d —————- (ii)
On subtracting equation (i) from (ii), we get,
– 10 = 2d
d = – 5
From equation (i), putting the value of d,we get
13 = a + (-5)
a = 18
a₃ = 18 + (3 – 1)(-5)
a₃ = 18 + 2(-5)
a₃ = 18 – 10
a₃ = 8
Therefore, The Series is 18, 13, 8, 3.

(iii) 5, $\square$ , $\square$ , $9\frac{1}{2}$
a = 5
a₄ = $9\frac{1}{2}$ = 19/2
As we know, for an A.P.,
a_n = 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
Therefore, The Series is 5, 13/2, 8, $9\frac{1}{2}$ .

(iv) -4, $\square$ , $\square$ , $\square$ , $\square$ , 6
a = −4
a₆ = 6
As we know, for an A.P.,
a_n = 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 Series is -4, -2, 0, 2, 4, 6.

(v) $\square$ , 38, $\square$ , $\square$ , $\square$ , -22
a₂ = 38
a₆ = −22
As we know, for an A.P.,
a_n = a + (n − 1)d
a₂ = a + (2 − 1)d
38 = a + d ————- (i)
a₆ = a + (6 − 1)d
−22 = a + 5d ————- (ii)
On subtracting equation (i) from (ii), we get
− 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 Series is 53, 38, 23, 8, -7, -22.

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

Solution – Given the A.P. series as 3, 8, 13, 18, …
First term, a = 3
Common difference, d = a₂ − a₁ = 8 − 3 = 5
a_n = 78
n = ?
a_n = a + (n − 1)d
78 = 3 + (n −1)5
75 = (n − 1)5
(n − 1) = 15
n = 16
Hence, 16^th 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, $\mathbf{15\frac{1}{2}}$ , 13, ……., -47

Solution –

(i) 7, 13, 19, …, 205
First term, a = 7
Common difference, d = a₂ − a₁ = 13 − 7 = 6
a_n = 205
n = ?
As we know, for an A.P.,
a_n = a + (n − 1) d
205 = 7 + (n − 1) 6
198 = (n − 1) 6
33 = (n − 1)
n = 34
So, there are 34 terms in the given A.P.

(ii) 18, $\mathbf{15\frac{1}{2}}$ , 13, ……., -47
First term, a = 18
Common difference, d = a₂– a₁= $15\frac{1}{2}$ – 18
d = (31 – 36)/2 = -5/2
a_n = -47
As we know, for an A.P.,
a_n = 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
Hence, there are 27 terms in the given A.P.

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

Solution – Given :
First term, a = 11
Common difference, d = a₂− a₁ = 8 − 11 = −3
Suppose -150 is the nth term of the given A.P. then it can be written as
a_n = a + (n − 1)d
-150 = 11 + (n – 1)(-3)
-150 = 11 – 3n + 3
-164 = -3n
n = 164/3
Clearly, n is not an integer but a fraction.
Therefore, – 150 is not a term of this A.P.

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

Solution – Given :
a₁₁ = 38
a₁₆ = 73
We know that,
a_n = a+(n−1)d
a₁₁ = a + (11 − 1)d
38 = a + 10d —————— (i)
a₁₆ = a + (16 − 1)d
73 = a + 15d —————— (ii)
On subtracting equation (i) from (ii), we get
15d – 10d = 73 – 38
5d = 35
d = 7
From equation (i), we can write,
38 = a + 10 × (7)
38 − 70 = a
a = −32
a₃₁ = a + (31 − 1) d
a₃₁ = − 32 + 30 (7)
a₃₁ = − 32 + 210
a₃₁ = 178
Hence, 31^st term of the given A.P. is 178.

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

Solution – Given :
a₃ = 12
a₅₀ = 106
a_n = a + (n − 1)d
a₃ = a + (3 − 1)d
12 = a + 2d —————- (i)
a₅₀= a + (50 − 1)d
106 = a + 49d —————- (ii)
On subtracting equation (i) from (ii), we get
(a + 49d) – (a + 2d) = 106 – 12
47d = 94
d = 2
From equation (i), we can write now,
12 = a + 2(2)
a = 12 − 4 = 8
a₂₉ = a + (29 − 1)d
a₂₉ = 8 + (28)2
a₂₉ = 8 + 56
a₂₉ = 64
So the 29^th term of the given A.P. will be 64.

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

Solution – Given :
a₃ = 4
a₉ = −8
a_n = 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 will get here,
−8 −4 = (a + 8d) – (a + 2d)
−12 = 6d
d = −2
From equation (i), we can write,
4 = a + 2(−2)
4 = a − 4
a = 8
Let n^th term of this A.P. be zero.
a_n= a + (n − 1)d
0 = 8 + (n − 1)(−2)
0 = 8 − 2n + 2
2n = 10
n = 5
Hence, the 5^th term of the given A.P. will be zero.

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

Solution – According to the question
a₁₇ = a₁₀ + 7
a + (17 – 1)d = a + (10 – 1)d + 7
(a + 16d) − (a + 9d) = 7
7d = 7
d = 1
Hence, the common difference of the given A.P. will be 1.

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

Solution – Given :
a = 3
d = a₂ − a₁ = 15 − 3 = 12
Let the nth term of the A.P. will be 132 more than its 54th term
a_n = a₅₄ + 132
a + (n − 1)d = a + (54 − 1)d + 132
(n − 1)12 = (53)(12) + 132
(n − 1)12 = 636 + 132
(n − 1)12 = 768
(n − 1) = 64
n = 64 + 1
n = 65
Hence, 65^th term will be 132 more than its 54^th term.

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

Solution – Let, the first term of two APs be a₁ and a₂ respectively
And the common difference of these APs be d.
For First A.P., we know,
a_n = a + (n − 1)d
a₁₀₀ = a₁+ (100 − 1)d
a₁₀₀ = a₁ + 99d
a₁₀₀₀ = a₁+ (1000 − 1)d
a₁₀₀₀ = a₁+ 999d

For second A.P., we know,
a_n = a + (n − 1)d
a₁₀₀ = a₂+ (100 − 1)d
a₁₀₀ = a₂+99d
a₁₀₀₀ = a₂+ (1000−1)d
a₁₀₀₀ = a₂+ 999d
According to the question, the difference between their 100th term is 100
(a₁ + 99d) – (a₂ + 99d) = 100
a₁− a₂ = 100 —————– (i)
Difference between 1000^th terms of the two APs
(a₁+ 999d) − (a₂+ 999d) = 100
From equation (i),
(a₁+ 999d) − (a₂+ 999d) = a₁− a₂Hence, the difference between their 1000^th term will be 100.

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

Solution – First three-digit number that is divisible by 7 are;
First number = 105
Second number = 105 + 7 = 112
Third number = 112 + 7 = 119
Therefore, 105, 112, 119, …
As we know, the largest possible three-digit number is 999.
When we divide 999 by 7, the remainder will be 5.
Therefore, 999 – 5 = 994 is the maximum possible three-digit number that is divisible by 7.
Now the series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
a = 105
d = 7
a_n = 994
n = ?
a_n = 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 – Given :
12 is the minimum number that is divisible by 4 between 10 and 250.
a = 12
d = 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, now;
12, 16, 20, 24, …, 248
a_n = 248
a_n = 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 – Given :
Two APs as; 63, 65, 67,… and 3, 10, 17,….
Taking First AP,
63, 65, 67, …
a = 63
d = a₂− a₁ = 65 − 63 = 2
a_n = a + (n − 1)d
a_n= 63 + (n − 1)2
a_n= 63 + 2n − 2
a_n = 61 + 2n ————— (i)

Taking second AP,
3, 10, 17, …
a = 3
d = a₂ − a₁ = 10 − 3 = 7
a_n = a + (n − 1)d
a_n = 3 + (n − 1)7
a_n = 3 + 7n − 7
a_n = 7n − 4 ————— (ii)
According to the question, n^th terms of both APs are equal
61 + 2n = 7n − 4
61 + 4 = 5n
5n = 65
n = 13
Therefore, 13^th terms of both these A.P.s are equal to each other.

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

Solutions – Given :
a₃ = 16
a + (3 − 1)d = 16
a + 2d = 16 —————- (i)
It is given that, 7^th term exceeds the 5^th term by 12.
a₇ − a₅ = 12
[a + (7 − 1)d] − [a + (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 be4, 10, 16, 22, …

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

Solution – Given :
d = 8 – 3 = 5
Therefore, we can write the given AP in reverse order as;
253, 248, 243, …, 13, 8, 5
Now for the new AP,
a = 253
d = 248 − 253 = −5
n = 20
Therefore, using nth term formula, we get,
a₂₀ = a + (20 − 1)d
a₂₀ = 253 + (19)(−5)
a₂₀ = 253 − 95
a = 158
Therefore, 20^th term from the last term of the AP 3, 8, 13, …, 253.is 158.

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

Solution – Given :
a₄ + a₈= 24
[a + (4 − 1)d] + [a + (8 − 1)d] = 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?

Solution – As, salary of Subba Rao is increasing by a fixed amount in every year hence this will form an AP with
First term (a) = 5000
common difference (d) = 200
a_n = 7000
a + (n – 1)d = 7000
5000 + (n – 1) 200 = 7000
n – 1 = 2000/200
n = 10 + 1
n = 11
Hence, the salary will be 7000 in 11^th year.

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 weekly savings become Rs 20.75, find n.

Solution – Given that, Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75.
First term, a = 5
Common difference, d = 1.75
a_n= 20.75
n = ?
a_n = a + (n − 1)d
20.75 = 5 + (n – 1)×1.75
15.75 = (n – 1)×1.75
(n – 1) = 15.75/1.75
(n – 1) = 1575/175
(n – 1) = 9
n – 1 = 9
n = 10
Hence, n is 10.

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NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2