# How many terms of the AP: $-15,-13,-11,-$ are needed to make the sum $-55$ ? Explain the reason for double answer.

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Given:

Given A.P. is $-15,-13,-11,......$

To do:

We have to find the number of terms that must be taken so that their sum is $-55$.

Solution:

Let the number of terms be $n$.

First term $(a)=-15$

Common difference $(d)=-13-(-15)=-13+15=2$

We know that,

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

$\Rightarrow -55=\frac{n}{2}[2 \times (-15)+(n-1) \times2]$

$\Rightarrow -55=\frac{n}{2}[-30+2n-2]$

$\Rightarrow -110=n(2n-32)$

$\Rightarrow 2n^{2}-32n+110=0)$

$\Rightarrow n^{2}-16 n+55=0$

$\Rightarrow n^{2}-11 n-5 n+55=0$

$\Rightarrow n(n-11)-5(n-11)=0$

$\Rightarrow(n-11)(n-5)=0$

This implies,

$n-11=0$ or $n-5=0$

$n=11$ or $n=5$

The number of terms needed to make the sum $-55$ is 5 or 11.

Updated on 10-Oct-2022 13:27:40