**CBSE NCERT Solutions Chapter 5 Arithmetic Progressions Exercise 5.3 Question 10**

**10. Show that form an AP where is defined as below:**

**(i) (ii) **

Also. find the sum of the first 15 terms in each case.

**Solution (i) **

We need to show that form an AP where

Lets calculate values of using .

So, the sequence is of the form

Lets check difference between consecutive terms of this sequence.

11- 7 = 4

15- 11 = 4

19 - 15 = 4

**Therefore, the difference between consecutive terms is constant which means terms form an AP.**

We have sequence

First term = a =7

Commom difference = d = 4

**Applying formula**, to find sum of n terms of AP , **we get**

**Therefore, sum of first 15 terms of AP is equal to 525.**

**Solution (ii) **

We need to show that form an AP where

Lets calculate values of using .

So, the sequence is of the form

Lets check difference between consecutive terms of this sequence.

-1-(4) = -5

-6-(-1) = -6 + 1 = -5

-11-(-6) = -11 + 6 = -5

**Therefore, the difference between consecutive terms is constant which means terms form an AP.**

We have sequence

First term = a =4

Commom difference = d = -5

**Applying formula**, to find sum of n terms of AP , **we get**

**Therefore, sum of first 15 terms of AP is equal to -465.**