Question

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 :

• A. 1862
• B. 1842
• C. 1852
• D. 1872

Answer 1

Answer: (A)

$S_{10}=530\Rightarrow \frac{10}{2}\{2a+9d\}=530$
$\Rightarrow 2a+9d=106\dots \dots .(1)$
and $S_5=140\Rightarrow \frac{5}{2}\{2a+4d\}=140$
$\Rightarrow 2a+4d=56\dots \dots (2)$
$\Rightarrow 5d=50\Rightarrow d=10\Rightarrow a=8$
Now, $S_{20}-S_6=\frac{20}{2}\{2a+19d\}-\frac{6}{2}\{2a+5d\}$
$=14a+175d$
$=(14\times 8)+(175\times 10)$
$=1862$