Let $S_n$ denote the sum of first $n$-terms of an arithmetic progression. If $S_{10}=530,{ S}_5=140,$ then $S_{20}-S_6$ is equal to:

We know that the sum of $n$ terms of an arithmetic progression with its first term as $a$ and common difference $d$ is $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right].$

Given, $S_{10}=530$

$\Rightarrow \frac{10}{2}\left[2a+9d\right]=530$

$\Rightarrow 2a+9d=106 \dots \left(1\right)$

And $S_5=140$

$\Rightarrow \frac{5}{2}\left[2a+4d\right]=140$

$\Rightarrow 2a+4d=56 \dots \left(2\right)$

Subtracting the two equation, we get, $5d=50$

$\Rightarrow d=10$

On putting the value of $d$ in the equation $\left(2\right),$ we get

$2a+4\times 10=56$

$\Rightarrow 2a=16$

$\Rightarrow a=8.$

Now, $S_{20}-S_6=\frac{20}{2}\left[2a+19d\right]-\frac{6}{2}\left[2a+5d\right]$

$=14a+175 d$

$=(14\times 8)+(175\times 10)$

$=1862.$