How many terms of the A.P. $27, 24, 21, …,$ should be taken so that their sum is zero?

Given:

Given A.P. is $27, 24, 21, …,$
To do:

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

Here,

First term $a=27$

Common difference $d=24-27=-3$

Let the number of terms to be taken be $n$.

We know that,

\( \mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d] \)
\( 0=\frac{n}{2}[(2 \times 27)+(n-1)(-3)] \)
\( 0=n[54-3 n+3] \)
\( 0=n[57-3 n] \)
$n=0$ which is not possible or \( (57-3 n)=0 \)
\( 57=3 n \)
\( \therefore n=\frac{57}{3}=19 \)

The number of terms to be taken is 19.   

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Updated on: 10-Oct-2022

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