Let S_{n} denote the sum of first n-terms of | vedclass

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Question

$S_{10}=530 \Rightarrow \frac{10}{2}\{2 a+9 d\}=530$

$\Rightarrow 2 a+9 d=106 \ldots .(1)$

$	ext { and } S_{5}=140 \Rightarrow \frac{5}{2}\{2 a

Vedclass · Accepted Answer

$1862$

Vedclass · Answer

$S_{10}=530 \Rightarrow \frac{10}{2}\{2 a+9 d\}=530$

$\Rightarrow 2 a+9 d=106 \ldots .(1)$

$	ext { 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, $S_{20}-S_{6}=\frac{20}{2}\{2 a+19 d\}-\frac{6}{2}\{2 a+5 d\}$

$=14 a+175 d$

$=(14 	imes 8)+(175 	imes 10)$

$=1862$

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 :

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