Mathematics - Sequence & Series Question with Solution | TestHub

Since,$H.C.F.$ of $\left(91, n\right)>1.$

Hence, $n$ should be multiple of $7$ or $13$ or both.

Hence, when $n$ is multiple of $7$,

Number of terms$=14,$ $t_1=105,$ $l=196$

$\therefore S_1=\frac{14}{2}\left(105+196\right)=7\times 301=2107$

When $n$ is multiple of $13$,

Number of terms $=8, t_1=104, l=195$

$\therefore S_2=\frac{8}{2}\times 299=1196$

When $n$ is multiple of $91$

$\therefore$ Number of term $=1$

$S_3=182$

$\therefore$ Required sum $=2107+1196-182=3121$