# How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

**Solution:**

Sum of the first n terms of an AP is given by S_{n} = n/2 [2a + (n - 1) d].

Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.

Given,

- First term, a = 9
- Common difference, d = 17 - 9 = 8
- Sum up to nth terms, S
_{n }= 636

We know that sum of n terms of AP

S_{n} = n [2a + (n - 1) d]

636 = n/2 [2 × 9 + (n - 1) 8]

636 = n/2 [18 + 8n - 8]

636 = n/2 [10 + 8n]

636 = n[5 + 4n]

636 = 5n + 4n²

4n² + 5n - 636 = 0

4n² + 53n - 48n - 636 = 0

n (4n + 53) - 12 (4n + 53) = 0

(4n + 53)(n - 12) = 0

Either 4n + 53 = 0 or n - 12 = 0

n = - 53/4 or n =12

n cannot be -53/4 because the number of terms can neither be negative nor fractional, therefore, n = 12

**Video Solution:**

## How many terms of the AP. 9, 17, 25 ... must be taken to give a sum of 636?

### Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.3 Question 4 - Chapter 5 Exercise 5.3 Question 4:

How many terms of the AP. 9, 17, 25 ... must be taken to give a sum of 636?

12 terms must be taken for AP. 9, 17, 25 ... to give a sum of 636