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How many terms of the A.P. 17, 15, 13, ...... must be taken, so that their sum is 81?

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answer is A.

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Detailed Solution

First term,

a_{1} = 17

Second term,

a_{2} = 15

Third term,

a_{3} = 13

Sum of n terms is 81,

S_{n} = 81

Common difference

d = a_{n} - a_{n - 1}

d = 15 - 17 = - 2

Now, use the formula of sum of n terms of AP,

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

81 = \frac{n}{2} [2 \times 17 + (n - 1) \times (- 2)]

162 = - 2 n^{2} + 36 n

n^{2} - 18 n + 81 = 0

(n - 9)^{2} = 0

n = 9

Thus, there are total 9 terms.
Correct option is 1.

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