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JEE Mains · Maths · STD 11 - 8. sequence and series
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 \(1872\)
- D \(1862\)
Answer & Solution
Correct Answer
(D) \(1862\)
Step-by-step Solution
Detailed explanation
\(S_{10}=530 \Rightarrow \frac{10}{2}\{2 a+9 d\}=530\) \(\Rightarrow 2 a+9 d=106 \ldots .(1)\) \(\text { and } S_{5}=140 \Rightarrow \frac{5}{2}\{2 a+4 d\}=140\) \(\Rightarrow 2 a+4 d=56 \ldots . .(2)\) \(\Rightarrow 5 d=50 \Rightarrow d=10 \Rightarrow a=8\) Now,…
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