# CBSE Past Year Board Paper

Solution to 2017 Class 10 Math - Set 3 Q19. Sum of AP

#### Video Explanation

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

First term a1 = 9
Common difference = 8
Sum upto ‘n’ terms = 636

Sum upto ‘n’ terms of an AP = $$frac{n}{2}$$2a1 + (n – 1)d)
636 = $$frac{n}{2}$$2 × 9 + (n – 1)8)
1272 = n(18 + 8n – 8)
1272 = 8n2 + 10n
4n2 + 5n – 636 = 0
4n2 + 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.

.