CBSE Past Year Board Paper

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

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 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.

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