Let S a denote the sum of first n terms an arithmetic progression. If S 20 = 790 and S 10 = 145 , then S 15…

Given,

In an arithmetic progression sum of first $20$ terms is $790$

So, let $a$ be the first term and $d$ be common difference,

So, $S_{20}=\frac{20}{2}\left[2a+19d\right]=790 .....\left(1\right)$

And sum of the first $10$ terms is $145$

So, $S_{10}=\frac{10}{2}\left[2a+9d\right]=145 .....\left(2\right)$

By solving equation $\left(1\right) \& \left(2\right)$ we get,

$a=-8 \& d=5$

 Then, $S_{15}-S_5=\frac{15}{2}\left[2\times \left(-8\right)+14\times 5\right]-\frac{5}{2}\left[2\times \left(-8\right)+4\times 5\right]$

$\Rightarrow S_{15}-S_5=405-10=395$