#### Question How many terms of an A.P. 9, 17, 25, .... must be taken to give a sum of 636?

#### Video Explanation

#### Explanatory Answer

The AP is 9, 17, 25, ....

First term a_{1} = 9

Common difference = 8

Sum upto ‘n’ terms = 636

Sum upto ‘n’ terms of an AP = \\frac{n}{2})(2a_{1} + (n – 1)d)

636 = \\frac{n}{2})(2 × 9 + (n – 1)8)

1272 = n(18 + 8n – 8)

1272 = 8n^{2} + 10n

4n^{2} + 5n – 636 = 0

4n^{2} + 53n – 48n – 636 = 0

n(4n + 53) – 12(4n + 53) = 0

(4n + 53) (n – 12) = 0

n = \\frac{-53}{4}), or n = 12

Because number of terms can neither be a fraction nor a negative number, n = 12.

12 terms are required to give a sum of 636.