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

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answer is A.

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Detailed Solution

Given series is,
the AP: 9, 17, 25…. In which,
First term,

a_{1} = 9

Second term a_{2} = a_{1} + d = 17

9 + d = 17

d = 8

Common difference, d = 8

Suppose total number of terms in an A.P. =n. Now, by the sum of n terms of AP formula,

S_{n} = \frac{n}{2} \{2 a + (n - 1) d\}

636 = \frac{n}{2} \{2 (9) + (n - 1) 8\}

1272 = 18 n + 8 n^{2} - 8 n

8 n^{2} + 10 n - 1272 = 0

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

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

Compare it with

{an}^{2} + bn + c = 0

and put the values in the quadratic formula,

n = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}

n = \frac{- 5 \pm \sqrt{5^{2} - 4 \cdot 4 (- 636)}}{2 \cdot 4}

n = \frac{- 5 \pm 101}{2 \cdot 4}

So,

n = \frac{- 5 + 101}{8}, n = \frac{- 5 - 101}{8}

n = 12, - 13.25

Do not consider the negative value.

n = 12

Thus, total terms are 12 terms to get the sum 636.
Option 1 is correct.

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