JEE Mains · Maths · STD 11 - 8. sequence and series
Let the sum of the first \(n\) terms of an A.P. be \(3n^2 + 5n\). Then the sum of squares of the first \(10\) terms of the A.P. is:
- A \(10220\)
- B \(12860\)
- C \(15220\)
- D \(19780\)
Answer & Solution
Correct Answer
(C) \(15220\)
Step-by-step Solution
Detailed explanation
Given the sum of the first \(n\) terms of the A.P. is \(S_n = 3n^2 + 5n\). The \(n\)-th term of the A.P. is given by: \(T_n = S_n - S_{n-1}\) \(T_n = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)]\) \(T_n = 3n^2 + 5n - [3(n^2 - 2n + 1) + 5n - 5]\)…
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