JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(S_{1}\) be the sum of first \(2 n\) terms of an arithmetic progression. Let, \(S_{2}\) be the sum of first \(4n\) terms of the same arithmetic progression. If \(\left( S _{2}- S _{1}\right)\) is \(1000,\) then the sum of the first \(6 n\) terms of the arithmetic progression is equal to:
- A \(1000\)
- B \(7000\)
- C \(5000\)
- D \(3000\)
Answer & Solution
Correct Answer
(D) \(3000\)
Step-by-step Solution
Detailed explanation
\(S_{2 n}=\frac{2 n}{2}[2 a+(2 n-1) d], S_{4 n}=\frac{4 n}{2}[2 a+(4 n-1)d]\) \(\Rightarrow S _{2}- S _{1}=\frac{4 n }{2}[2 a +(4 n -1) d ]-\frac{2 n }{2}[2 a +(2 n -1)d]\) \(=4 a n+(4 n-1) 2 n d-2 n a-(2 n-1) d n\) \(=2 n a+n d[8 n-2-2 n+1]\)…
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