In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.2. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 5 Arithmetic Progressions 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 a_{n} the n^{th} term of the A.P.

Solution:

Solution:

(i) a_{n} = a + (n- 1)d

a_{8} = 7 + (8 – 1)3 = 7 + 7 × 3 = 7 + 21

⇒ a_{8} = 28

(ii) a_{n} = a + (n – 1)d

⇒ a_{10} = -18 + (10 – 1)7 ⇒ 0 = -18 + 9d

⇒ 9d = 18 ⇒ d = \(\frac{18}{9}=2\)

∴ d = 2

(iii) a_{n} = a + (n – 1)d

⇒ -5 = a + (18 – 1) × (-3)

⇒ -5 = a + 17 × (-3)

⇒ -5 = a – 51 ⇒ a = -5 + 51 = 46

Thus, a = 46

(iv) a_{n} = a + (n – 1)d

⇒ 3.6 = -18.9 + (n – 1) × 2.5

⇒ (n – 1) × 2.5 = 3.6 + 18.9

⇒ n = 9 + 1 = 10

Thus, n = 10

(v) a_{n} = a + (n- 1)d

⇒ a_{n} = 3.5 + (105 – 1) × 0

⇒ a_{n} = 3.5 + 104 × 0 ⇒ a_{n} = 3.5 + 0 = 3.5

Thus, a_{n} = 3.5

Question 2.

Choose the correct choice in the following and justify:

(i) 30^{th} term of the AP: 10,7,4, , is, ….,

(A) 97

(B) 77

(C) -77

(D) -87

(ii) 11^{th} term of the AP: -3, \(-\frac{1}{2}\), 2, …. ,is

(A) 28

(B) 22

(C) -38

(D) -48\(\frac{1}{2}\)

Solution:

(i) (C): Here, a = 10, n = 30

∵ T_{10} = a + (n – 1)d and d = 7 – 10 = -3

∴ T_{30} = 10 + (30 – 1) × (-3)

⇒ T_{30} = 10 + 29 × (-3)

⇒ T_{30} = 10 – 87 = -77

(ii) (B): Here, a = -3, n = 11 and

Question 3.

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

Solution:

(i) Here, a = 2, T_{3} = 26

Let common difference = d

∴ T_{n} = a + (n- 1 )d

⇒ T_{3} = 2 + (3 – 1)d

⇒ 26 = 2 + 2 d

⇒ 2d = 26 – 2 = 24

(ii) Let the first term = a

and common difference = d

Here, T_{2} = 13 and T_{4} = 3

T_{2} = a + d = 13, T_{4} = a + 3d = 3

T_{1} – T_{2} = (a + 3d) – (a + d) = 3 – 13

(iii)

(iv) Here, a = – 4, T_{6} = 6

∵ T_{n} = a + (n -1 )d

T_{6} = – 4 + (6 – 1)d ⇒ 6 = -4 + 5d ⇒ 5d = 6 + 4 = d = 10 – 5 = 2

T_{2} = a + d = -4 + 2 =-2

T_{3} = a + 2d = -4 + 2(2) = 0

T_{4} = a + 3d = -4 + 3(2) = 2

T_{5} = a + 4d = -4 + 4(2) = 4

(v) Here, T_{2} = 38 and T_{6} = -22

∴ T_{2} = a + d = 38, T_{6} = a + 5d = -22

⇒ T_{6} – T_{2} = a + 5d – (a + d) = -22 – 38 -60

⇒ 4d = -60 ⇒ d = \(\frac{-60}{4}\) = -15

a + d = 38 ⇒ a + (-15) = 38

⇒ a = 38 + 15 = 53

Now,

T_{3} = a + 2d = 53 + 2(-15) = 53 – 30 = 23

T_{4} = a + 3d = 53 + 3(-15) = 53 – 45 = 8

T_{5} = a + 4d = 53 + 4(-15) = 53 – 60 = -7

Thus missing terms are

Question 4.

Which term of the AP: 3, 8, 13, 18, is 78?

Solution:

Let the n^{th} term be 78

Here, a = 3 ⇒ T_{1} = 3 and T_{2} = 8

∴ d = T_{2} – T_{1} = 8 – 3 = 5

And, T_{n} = a + (n- 1 )d

⇒ 78 = 3 + (n – 1) × 5 ⇒ 78 – 3 = (n -1) × 5

⇒ 75 = (n – 1) × 5 ⇒ (n – 1) = 75 ÷ 5 = 15

⇒ n = 15 + 1 = 16

Thus, 78 is the 16^{th} term of the given AP.

Question 5.

Find the number of terms in each of the following APs:

(i) 7,13,19, …….. ,205

(ii) 18, \(15 \frac{1}{2}\), 13, …… ,-47

Solution:

(i) Here, a = 7,d = 13 – 7 = 6

Let total number of terms be n.

∴ T_{n} = 205

Now, T_{n} = a + (n – 1) ×d

= 7 + (n – 1) × 6 = 205

⇒ (n – 1) × 6 = 205 – 7 = 198

⇒ n – 1 = \(\frac{198}{6}=33\)

∴ n = 33 + 1 = 34

Thus, the required number of terms is 34.

(ii)

Thus, the required number of terms is 27.

Question 6.

Check whether -150 is a term of the AP:

11, 8, 5, 2…

Solution:

For the given AP,

we have a = 11, d = 8 -11 = -3

Let -150 be the n^{th} term of the given AP

∴ T_{n} = a + (n – 1 )d

⇒ -150 = 11 + (n – 1) × (-3)

⇒ -150 – 11 = (n – 1) × (-3)

⇒ -161 = (n – 1) ⇒ (-3)

But n should be a positive integer.

Thus, -150 is not a term of the given AP

Question 7.

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

Solution:

Here, T_{11} = 38 and T_{16} = 73

Let the first term = a and the common difference = d.

T_{n} = a + (n – 1 )d

Then, T_{n} = a + (11 – 1)d = 38

⇒ a + 10d = 38 …(1)

and T_{16} = a + (16 – 1)d = 73

⇒ a + 15d = 73 …(2)

Subtracting (1) from (2), we get

(a + 15d) – (a + 10d) = 73 – 38

Question 8.

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

Solution:

Here, n = 50, T_{3} = 12, T_{n} = 106

⇒ T_{50} = 106

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

T_{n} = a + (n – 1 )d

T_{3} = a + 2d = 12 …(1)

T_{50} = a + 49d = 106 …(2)

Subtracting (1) from (2), we get

Question 9.

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

Solution:

Here, T_{3} = 4 and T_{9} = -8

T_{n} = a + (n – 1)d

T_{3} = a + 2d = 4 …. (1)

T_{9} = a + 8d = – 8 …. (2)

Subtracting (1) from (2), we get

Question 10.

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

Solution:

Let a be the first term and d the common difference of the given AP

Now, using _{n} = a + (n – 1 )d, we have

Thus, the common difference is 1.

Question 11.

Which term of the AP : 3, 15, 27, 39,… will be 132 more than its 54^{th} term?

Solution:

Here, a = 3, d = 15 – 3 = 12

Using T_{n} = a + (n – 1 )d, we get

Thus, 132 more than 54^{th} term is the 65^{th} term.

Question 12.

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

Solution:

Let for the 1^{st} AP, the first term = a and common difference = d

And for the 2^{nd} AP, the first term = a and common difference = d

According to the condition,

Question 13.

How many three-digit numbers are divisible by 7?

Solution:

The first three-digit number divisible by 7 is 105.

The last such three-digit number divisible by 7 is 994.

∴ The AP is 105,112,119, ,994

Let n be the required number of terms Here, a = 105, d = 7 and _{n} = 994

Thus, 128 numbers of 3-digits are divisible by 7.

Question 14.

How many multiples of 4 lie between 10 and 250?

Solution:

The first multiple of 4 beyond 10 is 12.

The multiple of 4 just below 250 is 248.

Thus, the required number of terms is 60.

Question 15.

For what value of n, are the n^{th} terms of two APs: 63,65,67 … and 3,10,17, …. equal?

Solution:

Thus, the 13^{th} terms of the two given AP’s are equal.

Question 16.

Determine the AP whose third term is 16 and the 7^{th} term exceeds the 5^{th} term by 12.

Solution:

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

Question 17.

Find the 20^{th} term from the last term of the AP : 3, 8, 13, …, 253.

Solution:

We have, the last term (l) = 253

Here, d = 8 – 3 = 5

Since, the n^{th} term before the last term is given by l – (n – 1 )d

We have 20^{th} term from the last term = l – (20 – 1) × 5 = 253 – 19 × 5 = 253 – 95 = 158

Question 18.

The sum of the 4^{th} and 8^{th} 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.

Solution:

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

Question 19.

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

Solution:

Here, a = ₹ 5000 and d = ₹ 200

Let, in the nth year Subba Rao gets ₹ 7000.

Question 20.

Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the n^{th} week, her weekly savings become ₹ 20.75, find n.

Solution:

Here, a = ₹ 5 and d = ₹ 1.75

Thus, the required number of weeks is 10.

In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.2. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 5 Arithmetic Progressions 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 a_{n} the n^{th} term of the A.P.

Solution:

Solution:

(i) a_{n} = a + (n- 1)d

a_{8} = 7 + (8 – 1)3 = 7 + 7 × 3 = 7 + 21

⇒ a_{8} = 28

(ii) a_{n} = a + (n – 1)d

⇒ a_{10} = -18 + (10 – 1)7 ⇒ 0 = -18 + 9d

⇒ 9d = 18 ⇒ d = \(\frac{18}{9}=2\)

∴ d = 2

(iii) a_{n} = a + (n – 1)d

⇒ -5 = a + (18 – 1) × (-3)

⇒ -5 = a + 17 × (-3)

⇒ -5 = a – 51 ⇒ a = -5 + 51 = 46

Thus, a = 46

(iv) a_{n} = a + (n – 1)d

⇒ 3.6 = -18.9 + (n – 1) × 2.5

⇒ (n – 1) × 2.5 = 3.6 + 18.9

⇒ n = 9 + 1 = 10

Thus, n = 10

(v) a_{n} = a + (n- 1)d

⇒ a_{n} = 3.5 + (105 – 1) × 0

⇒ a_{n} = 3.5 + 104 × 0 ⇒ a_{n} = 3.5 + 0 = 3.5

Thus, a_{n} = 3.5

Question 2.

Choose the correct choice in the following and justify:

(i) 30^{th} term of the AP: 10,7,4, , is, ….,

(A) 97

(B) 77

(C) -77

(D) -87

(ii) 11^{th} term of the AP: -3, \(-\frac{1}{2}\), 2, …. ,is

(A) 28

(B) 22

(C) -38

(D) -48\(\frac{1}{2}\)

Solution:

(i) (C): Here, a = 10, n = 30

∵ T_{10} = a + (n – 1)d and d = 7 – 10 = -3

∴ T_{30} = 10 + (30 – 1) × (-3)

⇒ T_{30} = 10 + 29 × (-3)

⇒ T_{30} = 10 – 87 = -77

(ii) (B): Here, a = -3, n = 11 and

Question 3.

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

Solution:

(i) Here, a = 2, T_{3} = 26

Let common difference = d

∴ T_{n} = a + (n- 1 )d

⇒ T_{3} = 2 + (3 – 1)d

⇒ 26 = 2 + 2 d

⇒ 2d = 26 – 2 = 24

(ii) Let the first term = a

and common difference = d

Here, T_{2} = 13 and T_{4} = 3

T_{2} = a + d = 13, T_{4} = a + 3d = 3

T_{1} – T_{2} = (a + 3d) – (a + d) = 3 – 13

(iii)

(iv) Here, a = – 4, T_{6} = 6

∵ T_{n} = a + (n -1 )d

T_{6} = – 4 + (6 – 1)d ⇒ 6 = -4 + 5d ⇒ 5d = 6 + 4 = d = 10 – 5 = 2

T_{2} = a + d = -4 + 2 =-2

T_{3} = a + 2d = -4 + 2(2) = 0

T_{4} = a + 3d = -4 + 3(2) = 2

T_{5} = a + 4d = -4 + 4(2) = 4

(v) Here, T_{2} = 38 and T_{6} = -22

∴ T_{2} = a + d = 38, T_{6} = a + 5d = -22

⇒ T_{6} – T_{2} = a + 5d – (a + d) = -22 – 38 -60

⇒ 4d = -60 ⇒ d = \(\frac{-60}{4}\) = -15

a + d = 38 ⇒ a + (-15) = 38

⇒ a = 38 + 15 = 53

Now,

T_{3} = a + 2d = 53 + 2(-15) = 53 – 30 = 23

T_{4} = a + 3d = 53 + 3(-15) = 53 – 45 = 8

T_{5} = a + 4d = 53 + 4(-15) = 53 – 60 = -7

Thus missing terms are

Question 4.

Which term of the AP: 3, 8, 13, 18, is 78?

Solution:

Let the n^{th} term be 78

Here, a = 3 ⇒ T_{1} = 3 and T_{2} = 8

∴ d = T_{2} – T_{1} = 8 – 3 = 5

And, T_{n} = a + (n- 1 )d

⇒ 78 = 3 + (n – 1) × 5 ⇒ 78 – 3 = (n -1) × 5

⇒ 75 = (n – 1) × 5 ⇒ (n – 1) = 75 ÷ 5 = 15

⇒ n = 15 + 1 = 16

Thus, 78 is the 16^{th} term of the given AP.

Question 5.

Find the number of terms in each of the following APs:

(i) 7,13,19, …….. ,205

(ii) 18, \(15 \frac{1}{2}\), 13, …… ,-47

Solution:

(i) Here, a = 7,d = 13 – 7 = 6

Let total number of terms be n.

∴ T_{n} = 205

Now, T_{n} = a + (n – 1) ×d

= 7 + (n – 1) × 6 = 205

⇒ (n – 1) × 6 = 205 – 7 = 198

⇒ n – 1 = \(\frac{198}{6}=33\)

∴ n = 33 + 1 = 34

Thus, the required number of terms is 34.

(ii)

Thus, the required number of terms is 27.

Question 6.

Check whether -150 is a term of the AP:

11, 8, 5, 2…

Solution:

For the given AP,

we have a = 11, d = 8 -11 = -3

Let -150 be the n^{th} term of the given AP

∴ T_{n} = a + (n – 1 )d

⇒ -150 = 11 + (n – 1) × (-3)

⇒ -150 – 11 = (n – 1) × (-3)

⇒ -161 = (n – 1) ⇒ (-3)

But n should be a positive integer.

Thus, -150 is not a term of the given AP

Question 7.

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

Solution:

Here, T_{11} = 38 and T_{16} = 73

Let the first term = a and the common difference = d.

T_{n} = a + (n – 1 )d

Then, T_{n} = a + (11 – 1)d = 38

⇒ a + 10d = 38 …(1)

and T_{16} = a + (16 – 1)d = 73

⇒ a + 15d = 73 …(2)

Subtracting (1) from (2), we get

(a + 15d) – (a + 10d) = 73 – 38

Question 8.

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

Solution:

Here, n = 50, T_{3} = 12, T_{n} = 106

⇒ T_{50} = 106

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

T_{n} = a + (n – 1 )d

T_{3} = a + 2d = 12 …(1)

T_{50} = a + 49d = 106 …(2)

Subtracting (1) from (2), we get

Question 9.

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

Solution:

Here, T_{3} = 4 and T_{9} = -8

T_{n} = a + (n – 1)d

T_{3} = a + 2d = 4 …. (1)

T_{9} = a + 8d = – 8 …. (2)

Subtracting (1) from (2), we get

Question 10.

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

Solution:

Let a be the first term and d the common difference of the given AP

Now, using _{n} = a + (n – 1 )d, we have

Thus, the common difference is 1.

Question 11.

Which term of the AP : 3, 15, 27, 39,… will be 132 more than its 54^{th} term?

Solution:

Here, a = 3, d = 15 – 3 = 12

Using T_{n} = a + (n – 1 )d, we get

Thus, 132 more than 54^{th} term is the 65^{th} term.

Question 12.

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

Solution:

Let for the 1^{st} AP, the first term = a and common difference = d

And for the 2^{nd} AP, the first term = a and common difference = d

According to the condition,

Question 13.

How many three-digit numbers are divisible by 7?

Solution:

The first three-digit number divisible by 7 is 105.

The last such three-digit number divisible by 7 is 994.

∴ The AP is 105,112,119, ,994

Let n be the required number of terms Here, a = 105, d = 7 and _{n} = 994

Thus, 128 numbers of 3-digits are divisible by 7.

Question 14.

How many multiples of 4 lie between 10 and 250?

Solution:

The first multiple of 4 beyond 10 is 12.

The multiple of 4 just below 250 is 248.

Thus, the required number of terms is 60.

Question 15.

For what value of n, are the n^{th} terms of two APs: 63,65,67 … and 3,10,17, …. equal?

Solution:

Thus, the 13^{th} terms of the two given AP’s are equal.

Question 16.

Determine the AP whose third term is 16 and the 7^{th} term exceeds the 5^{th} term by 12.

Solution:

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

Question 17.

Find the 20^{th} term from the last term of the AP : 3, 8, 13, …, 253.

Solution:

We have, the last term (l) = 253

Here, d = 8 – 3 = 5

Since, the n^{th} term before the last term is given by l – (n – 1 )d

We have 20^{th} term from the last term = l – (20 – 1) × 5 = 253 – 19 × 5 = 253 – 95 = 158

Question 18.

The sum of the 4^{th} and 8^{th} 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.

Solution:

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

Question 19.

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

Solution:

Here, a = ₹ 5000 and d = ₹ 200

Let, in the nth year Subba Rao gets ₹ 7000.

Question 20.

Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the n^{th} week, her weekly savings become ₹ 20.75, find n.

Solution:

Here, a = ₹ 5 and d = ₹ 1.75

Thus, the required number of weeks is 10.