Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Arithmetic Progressions:

Exercise 5.1

Multiple Choice Questions (Q. No. 9-18)

Question. 10 What is the common difference of an AP in which a₁₈ − a₁₄ = 32?

(a) 8

(b) – 8

(c) – 4

(d) 4

Solution. (a)

Explanation:

Given, a₁₈ - a₁₄ = 32

Using a_n = a + (n – 1)d, we have:

            a₁₈ =  a + (18 − 1) d

And     a₁₃ = a + (13 − 1) d

Thus, a₁₈ – a₁₃ = [a + (18 − 1)d] − [a + (14 − 1) d] = 32

⟹       a + 17d – a – 13d = 32

⟹       4d = 32

⟹       d = 8

Thus, the required common difference of given AP = 8.

Question. 11 Two APs have the same common difference. The first term of one of these is − 1 and that of the other is − 8. The difference between their 4th terms is

(a) - 1

(b) – 8

(c) 7

(d) - 9

Solution. (c)

Explanation:

Let the same common difference of two APs be d.

Given, first term of first AP (a₁) = − 1

and the first term of second AP (a₁’) = − 8

We know that, the nth term of an AP is given as,

            a_n = a + (n − 1) d

Therefore, 4th term of first AP is,

            a₄ = a₁ + (4 − 1) d = −1 + 3d

And 4th term of second AP is,

            a₄’ = a₁’ + (4 − 1) d = − 8 + 3d

Difference between 4th terms of both APs is:

            a₄ − a₄’ = (−1 + 3d) − (− 8 + 3d)

                        = − 1 + 3d + 8 − 3d = 7

Hence, the required difference is 7.

Question. 12 If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be

(a) 7

(b) 11

(c) 18

(d) 0

Solution. (d)

Explanation:

We know that, the nth term of an AP is given as,

            a_n = a + (n − 1) d

Now, according to the question,

                                 7a₇ = 11a₁₁

⟹       7 [a + (7 − 1) d] = 11 [a + (11-1) d]                           

⟹                  7 (a +6d) = 11 (a + 10d)

⟹                  7a + 42d = 11a + 110d

⟹                   4a + 68d = 0

⟹                4 (a +17d) = 0

⟹                       a + 7d = 0                            [4 ≠ 0]

Or              a + (18 - 1) d = 0  

⟹18^th term of an AP, a₁₈ = 0

Question. 13 The 4th term from the end of an AP − 11, − 8, − 5,..., 49  is

(a) 37

(b) 40

(c) 43

(d) 58

Solution. (b)

Explanation:

Taking the AP in reverse order: 49,…, −5, −8, −11.

Here, a_l = 49   

Common difference, d = − 8 − (− 11)

                                       = − 8 + 11= 3

Now, we know that, the nth term of an AP is given as:

            a_n = a_l – (n − 1) d       

Therefore, fourth term of AP is,

            a₄ = 49 − (4 − 1) 3 = 49 − 9 = 40

Question. 14 The famous mathematician associated with finding the sum of the first 100 natural numbers is

(a) Pythagoras

(b) Newton

(c) Gauss

(d) Euclid

Sol. (c)

Explanation:

Gauss is the famous mathematician associated with finding the sum of the first 100 natural numbers i.e., 1 + 2 + 3 +...+ 100.

Question. 15 If the first term of an AP is − 5 and the common difference is 2, then the sum of the first 6 terms is

(a) 0

(b) 5

(c) 6

(d) 15

Solution. (a)

Explanation:

Question. 16 The sum of first 16 terms of the AP 10, 6, 2,… is

(a) −320

(b) 320

(c) −352

(d) −400

Solution. (a)

Explanation:

Question. 17 In an AP, if a = 1, a_n = 20 and S_n = 399, then n is equal to

(a) 19

(b) 21

(c) 38

(d) 42

Solution. (c)

Explanation:

Question. 18 The sum of first five multiples of 3 is

(a) 45

(b) 55

(c) 65

(d) 75

Solution. (a)

Explanation:

The first five multiples of 3 are 3, 6, 9, 12 and 15.

3, 6, 9, 12 and 15 form an AP with both its first term and common difference (6 – 3) being 3.

i.e., a = 3 and d = 3

Also, number of terms, n = 5