# 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, a_{k+1} – a_{k} 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.