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JEE Mains · Maths · STD 11 - 8. sequence and series

Given sum of the first \(n\) terms of an \(A.P.\) is \(2n + 3n^2.\) Another \(A.P.\) is formed with the same first term and double of the common difference, the sum of \(n\) terms of the new \(A.P.\) is

  1. A \(n + 4n^2\)
  2. B \(6n^2 - n\)
  3. C \(n^2 + 4n\)
  4. D \(3n + 2n^2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(6n^2 - n\)

Step-by-step Solution

Detailed explanation

Given \({S_n} = 2n + 3{n^2}\) Now, first term \(=2+3=5\) second term \(2(2)+3(4)=16\) third term \(=2(3)+3(9)=33\) Now, sum given in option \((b)\) only has the same first term and difference between \({2^{nd}}\) and \({1^{st}}\) term is double also.
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