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Sequence and Series

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Question : 1 of 86

Marks: +1, -0

The sum of a certain number of terms of an AP series

-8,-6,-4, . . . . .

52 .

The number of terms is

12
13
11
None

Solution:

Concept:

Sum of

n

terms of an AP:

S_n = \frac{n}{2}[2a + (n-1)d]

Explanation:

Given AP:

-8, -6, -4, \ldots

First term

a = -8

, common difference

d = 2

, sum

S_n = 52

S_n = \frac{n}{2}[2(-8) + (n-1)2] = \frac{n}{2}[-16 + 2n - 2] = \frac{n}{2}(2n - 18) = n(n-9)

Set

n(n-9) = 52

\Rightarrow n^2 - 9n - 52 = 0

Solve:

n = \frac{9 \pm \sqrt{81+208}}{2} = \frac{9 \pm \sqrt{289}}{2} = \frac{9 \pm 17}{2}

n = 13

(positive) or

n = -4

(invalid)

Answer:

13

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