# 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] or S_{n} = n/2 [a + l]

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

Given,

- First term, a = 9
- Common difference, d = 17 - 9 = 25 - 17 = ... = 8
- Sum up to n
^{th}terms, S_{n}= 636

We know that sum of n terms of an AP

S_{n} = n/2 [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 + 4n^{2}

4n^{2} + 5n - 636 = 0

4n^{2} + 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

**☛ Check:** NCERT Solutions for Class 10 Maths Chapter 5

**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

**Summary:**

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

**☛ Related Questions:**

- The first and the last term of an AP are 17 and 350 respectively.If the common difference is 9,how many terms are there and what is their sum?
- Find the sum of the first 22 terms of an AP in which d = 7 and 22nd term is 149.
- Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
- If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

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