Mathematics - Sequence & Series Question with Solution | TestHub

Let the first term of$A.P.$is$‘a’$and the common difference is$‘d’$.
$\therefore S_4=\frac{4}{2}\left(2a+\left(4-1\right)d\right)=16$
$\Rightarrow 2a+3d=8 ...\left(i\right)$
and$S_6=\frac{6}{2}\left(2a+\left(6-1\right)d\right)=-48$
$\Rightarrow 2a+5d=-16 ...\left(ii\right)$
$\therefore$From$\left(i\right) \& \left(ii\right)$,$a=22$$\&$$d=-12$
$\therefore S_{10}=\frac{10}{2}(2a+9d)=5\left(2\times 22+9\left(-12\right)\right)$
$=-320$