Class 10 Maths

# Arithmetic Progression

## NCERT Exercise 5.1

### Part 3

Question: 4 – Which of the following are APs? If they form an AP, find the common difference and write three more terms.

(i) 2, 4, 8, 16, ………….

Solution: We know that, if a_(k+1) – a_k is same for different values of k then the series is an AP.

We have, a_1 = 2, a_2 = 4, a_3 = 8 and a_4 = 16

a_4 – a_3 = 16 – 8 = 8

a_3 – a_2 = 8 – 4 = 5

a_2 – a_1 = 4 – 2 = 2

Here, a_(k+1) – a_k is not same for all values of k.

Hence, the given series is not an AP.

(ii) 2, 5/2, 3, 7/2

Solution: We have, a_1 = 2, a_2 = 5/2, a_3 = 3 and a_4 = 7/2

a_4-a_3=7/2-3=1/2

a_3-a_2=3-5/2=1/2

a_2-a_1=5/2-2=1/2

Here, ak+1 – ak is same for all values of k.

Hence, the given series is an AP.

The common difference = ½

Next three terms of this series can be calculated as follows:

a_5=a+4d

=2+4xx1/2=4

a_6=a+5d

=2+5xx1/2=9/2

a_7=a+6d

=2+6xx1/2=5

Next three terms of the AP are: 4, 9/2 and 5

(iii) – 1.2, - 3.2, - 5.2, - 7.2, …….

Solution: a_4 – a_3 = - 7.2 + 5.2 = - 2

a_3 – a_2 = - 5.2 + 3.2 = - 2

a_2 – a_1 = - 3.2 + 1.2 = - 2

Here, a_(k+1) – a_k is same for all values of k.

Hence, the given series is an AP.

Common difference = - 2

Next three terms of the series can be calculated as follows:

a_5 = a + 4d

= -1.2 + 4 xx (-2)

= -1.2 - 8 = -9.2

a_6 = a + 5d

= -1.2 + 5 xx (-2)

= -1.2 - 10 = -11.2

a_7 = a + 6d

= -1.2 + 6 xx (-2)

= -1.2 - 12 = -13.2

Next three terms of AP are: - 9.2, - 11.2 and – 13.2

(iv) – 10, - 6, - 2, 2, …….

Solution: a_4 – a_3 = 2 + 2 = 4

a_3 – a_2 = - 2 + 6 = 4

a_2 – a_1 = - 6 + 10 = 5

Here, a_(k+1) – a_k is same for all values of k

Hence, the given series is an AP.

Common difference = 4

Next three terms of the AP can be calculated as follows:

a_5 = a + 4d

= -10 + 4 xx 4

= -10 + 16 = 6

a_6 = a + 5d

= -10 + 5 × 4

= -10 + 20 = 10

a_7 = a + 6d

= -10 + 6 xx 4

= -10 + 24 = 14

Next three terms of AP are: 6, 10 and 14.

(v) 3, 3 + sqrt2, 3 + 2sqrt2, 3 + 3sqrt2, ……

Solution: a_4 – a_3 = 3 + 3sqrt2 – 3 - 2sqrt2 = sqrt2

a_3 – a_2 = 3 + 2sqrt2 – 3 - sqrt2 = sqrt2

a_2 – a_1 = 3 + sqrt2 – 3 = sqrt2

Here, a_(k+1) – a_k is same for all values of k

Hence, the given series is an AP.

Common difference = sqrt2

Next three terms of the AP can be calculated as follows:

a_5 = a + 4d = 3 + 4sqrt2

a_6 = a + 5d = 3 + 5sqrt2

a_7 = a + 6d = 3 + 6sqrt2

Next three terms of AP are: 3 + 4sqrt2, 3 + 5sqrt2 and 3 + 6sqrt2

(vi) 0.2, 0.22, 0.222, 0.2222, ……

Solution: a_4 – a_3 = 0.2222 – 0.222 = 0.0002

a_3 – a_2 = 0.222 – 0.22 = 0.002

a_2 – a_1 = 0.22 – 0.2 = 0.02

Here, a_(k+1) – a_k is not same for all values of k

Hence, the given series is not an AP.