Mathematics - Sequence & Series Question with Solution | TestHub

Given sum of $n$ terms is $S_n=50n+\frac{n\left(n-7\right)}{2}A \dots \left(1\right)$

$\Rightarrow S_{n-1}=50(n-1)+\frac{(n-1)\left(n-8\right)}{2}A \dots \left(2\right)$

Subtracting $\left(1\right)$ and $\left(2\right),$ we get

$S_n-S_{n-1}=50n-50\left(n-1\right)+\frac{n\left(n-7\right)}{2}A-\frac{\left(n-1\right)\left(n-8\right)}{2}A$

$\Rightarrow S_n-S_{n-1}=50\left(n-n+1\right)+\frac{A}{2}\left(n\left(n-7\right)-\left(n-1\right)\left(n-8\right)\right)$

$\Rightarrow S_n-S_{n-1}=50+\frac{A}{2}\left(n^2-7n-n^2+9n-8\right)$

$\Rightarrow T_n=S_n-S_{n-1}=50+A(n-4)$

Hence, $T_1=50-3A$ and $T_2=50-2A$

$\Rightarrow d=T_2-T_1=A$ and $T_{50}=50+46A.$