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: Sequence & Series (Mathematics)

$\begin{aligned}& S_{10}=530 \Rightarrow \frac{10}{2}\{2 a+9 d\}=530 \\& \Rightarrow 2 a+9 d=106 \ldots(1) \\& \text { and } S_5=140 \Rightarrow \frac{5}{2}\{2 a+4 d\}=140 \\& \Rightarrow 2 a+4 d=56 \ldots(2) \\& \Rightarrow 5 d=50 \Rightarrow d=10 \Rightarrow a=8 \\& \text { Now, } S_m-S_b=\frac{20}{2}\{2 a+19 d\}-\frac{6}{2}\{2 a+5 d\} \\& =14 a+175 d \\& =(14 \times 8)+(175 \times 10) \\& =1862\end{aligned}$

Explanation

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