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

Exercise 5.1

Multiple Choice Questions (Q. No. 1-9)

Question. 1 In an AP, if d = − 4, n = 7 and a_n = 4, then a is equal to

(a) 6

(b) 7

(c) 20

(d) 28

Solution. (d)

Explanation:

We know that in an AP,

     a_n = a + (n − 1) d

⟹ 4 = a + (7 − 1) (− 4)                       [by given conditions]

⟹ 4 = a + 6 (− 4)

⟹ a = 4 + 24 = 28

Question. 2 In an AP, if a = 3.5, d = 0 and n = 101, then a_n will be

(a) 0

(b) 3.5

(c) 103.5

(d) 104.5

Solution. (b)

Explanation:

We know that in an AP,

      a_n = a + (n − 1) d

⟹ a_n = 3.5 + (101 − 1) × 0     [by given conditions]

⟹ a_n = 3.5

Question. 3 The list of numbers − 10, − 6, − 2, 2, ... is

(a) an AP with d = −16

(b) an AP with d = 4

(c) an AP with d = − 4

(d) not an AP

Solution. (b)

Explanation:

Given list of numbers is: −10, − 6, − 2, 2,....

Here, a₁ = −10, a₂ = − 6, a₃ = − 2 and a₄ = 2,

Now, d₁ = a₂ – a₁ = − 6 − (−10) = − 6 + 10 = 4

d₂ = a₃ – a₁ = − 2 − (− 6) = − 2 + 6 = 4

And d₃ =  a₄ - a₃ = 2 − (− 2) = 2 + 2 = 4

As d₁ = d₂ = d₃ = 4

So, the given list forms an AP with common difference, d = 4.

(a) –20

(b) 20

(c) –30

(d) 30

Solution. (b)

Explanation:

Question. 5 The first four terms of an AP whose first term is − 2 and the common difference is − 2 are

(a) − 2, 0, 2, 4

(b) − 2, 4, − 8, 16

(c) − 2, − 4, − 6, − 8

(d) − 2, − 4, − 8, − 16

Solution. (c)

Explanation:

Given, a₁ = –2 and  d = –2

Therefore, a₂ = a₁ + d

⟹       a₂ = –2 – 2 = – 4

Also     a₃ = a₂ + d = – 4 + (–2) = –6

And     a₄ = a₃ + d = – 6 + (–2) = –8

So, the first four terms are –2, –4, –6, –8.

Question. 6 The 21st term of an AP whose first two terms are − 3 and 4, is

(a) 17

(b) 137

(c) 143

(d) −143

Solution. (b)

Explanation:

For an AP, a_n = a + (n – 1)d  …..(i)

Given, first two terms of an AP are − 3 and a + d = 4.

⟹  a = − 3 and a₁= −3 + d = 4

⟹ Common difference, d = a1 – a = 4 – (−3) = 4+3 = 7

Therefore, a₂₁ = a + (21 − 1) d            [Using (i)]

                       = − 3 + (20) 7

                       = − 3 + 140 = 137

Question. 7 If the 2nd term of an AP is 13 and 5th term is 25, what is its 7th term?

(a) 30

(b) 33

(c) 37

(d) 38

Solution. (b)

Explanation:

Given, a₂ = 13 and a₅ = 25

Using a_n = a + (n – 1)d, we have:

      a + (2 − 1) d = 13 and      a + (5 − 1) d = 25

⟹  a + d = 13 ….(i)  and a + 4 d = 25   …..(ii)

On subtracting equation (i) from equation (ii), we get

     3 d = 25 – 13 = 12

⟹  d = 4

Putting value of d in equation (i), we get:

       a = 13 − 4 = 9

⟹  a₇ = a + (7 − 1) d = 9 + 6 × 4 = 33

Question. 8 Which term of an AP : 21, 42, 63, 84, ... is 210?

(a) 9^th

(b) 10^th

(c) 11^th

(d) 12^th

Solution. (b)

Explanation:

Let nth term of the given AP be 210.

Here, first term, a = 21

Common difference, d = 42 − 21 = 21

And a_n = 210

Using a_n = a + (n – 1)d, we have:

       210 = 21 + (n - 1) 21

⟹   210 = 21+ 21n – 21

⟹   210 = 21 n

⟹      n = 10

Hence, the 10th term of an AP is 210.

Question. 9 If the common difference of an AP is 5, then what is a₁₈ - a₁₃?

(a) 5

(b) 20

(c) 25

(d) 30

Solution. (c)

Explanation:

Given, common difference of an AP, d = 5

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 + (13 -1) d]                   

                        = a + 17 × 5 – a – 12 × 5

                        = 85 – 60 = 25