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How many terms of the sequence $18, 16, 14, . . . .$ should be taken so that their sum is zero?

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19

17

18

16

answer is A.

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Detailed Solution

Given a sequence

18, 16, 14, . . . .

We have to find how many terms of the sequence

18, 16, 14, . . . .

should be taken so that their sum becomes zero.
Common difference,

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

\Rightarrow

d = 16 - 18

\Rightarrow d = - 2

Here, we have

a = 18, d = - 2

.
Let there be

n

terms so that the sum becomes zero.
Now we have,

S_{n} = \frac{n}{2} [2 a + (n - 1) d] = 0

\Rightarrow \frac{n}{2}

[2 \times 18 + (n - 1) \times - 2] = 0

\Rightarrow [2 \times 18 + (n - 1) \times - 2] = 0

\Rightarrow 36 + (n - 1) \times - 2 = 0

\Rightarrow (n - 1) \times - 2 =

-

\Rightarrow n - 1 = \frac{- 36}{- 2}

\Rightarrow n - 1 = 18

\Rightarrow n = 18 + 1

\Rightarrow n = 19

As a result, the sequence's first 19 terms must be taken in order for their sum to be zero.
Hence, option 1 is correct.

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