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

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.