Question

Let $S_n$ denotes the sum of the first $n$ terms of an A.P.. If $S_4=16$ and $S_6=-48,$ then $S_{10}$ is equal to

• A. $-260$
• B. $-320$
• C. $-380$
• D. $-410$

Answer 1

The correct option is B $-320$

Let $a$ be the first term and $d$ is common difference of the A.P.
Given, $S_4=\frac{4}{2}(2a+3d)=16$
$\Rightarrow 2a+3d=8\dots \left(1\right)$

and $S_6=\frac{6}{2}(2a+5d)=-48$
$\Rightarrow 2a+5d=-16\dots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get
$a=22$ and $d=-12$

$\therefore S_{10}=\frac{10}{2}(2a+9d)=10\times 22+45\times (-12)$
$\therefore S_{10}=-320$


$\underset{–}{\text{Alternate Approach}}$
$S_6-S_4=a_5+a_6=-48-16=-64$

If there are $10$ terms in an A.P., then the mean of the $10$ terms is given by$=\frac{a_5+a_6}{2}=-32$
Therefore, Sum $S_{10}=-32\times 10=-320$