MCQ
Let $S_{n}$ denote the sum of first $n$-terms of an arithmetic progression. If $S_{10}=530, S_{5}=140$, then $\mathrm{S}_{20}-\mathrm{S}_{6}$ is equal to :
- A$1852$
- B$1842$
- C$1872$
- ✓$1862$
$\Rightarrow 2 a+9 d=106 \ldots .(1)$
$\text { and } S_{5}=140 \Rightarrow \frac{5}{2}\{2 a+4 d\}=140$
$\Rightarrow 2 a+4 d=56 \ldots . .(2)$
$\Rightarrow 5 d=50 \Rightarrow d=10 \Rightarrow a=8$
Now, $S_{20}-S_{6}=\frac{20}{2}\{2 a+19 d\}-\frac{6}{2}\{2 a+5 d\}$
$=14 a+175 d$
$=(14 \times 8)+(175 \times 10)$
$=1862$
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