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

\ S_10=530 => (10 / 2)\2 a+9 d\=530 => 2 a+9 d=106 ... (1) and S_5=140 => (5 / 2)\2 a+4 d\=140 => 2 a+4 d=56 ... (2) => 5 d=50 => d=10 => a=8 Now, S_m-S_b=(20 / 2)\2 a+19 d\-(6 / 2)\2 a+5 d\ =14 a+175 d =(14 x 8)+(175 x 10) =1862\

Let S_{n} denote the sum of first n-terms of an arithme... | 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

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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