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

Given:

The given sequence is $18, 16, 14,.…$

To do:

We have to find the number of terms that should be taken so that their sum is zero.

Solution:

Here,

First term \( (a)=18 \) and common difference \( (d)=16-18=-2 \)

Let \( n \) be the number of terms

We know that,

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 0=\frac{n}{2}[2 \times 18+(n-1)(-2)]$

$\Rightarrow 0=n[36-2 n+2]$

$\Rightarrow 0=n(38-2 n)$

$\Rightarrow n=0$, which is not possible or \( 38-2 n=0 \)

\( \Rightarrow 2 n=38 \)

$\Rightarrow n=19$

The number of terms in the sequence is 19. 

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Updated on: 10-Oct-2022

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