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)

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