Sequences And Series — Hard JEE math MCQ
If the sum of the first \(n\) terms of an AP is given by \(S_n = 3n^2 + 5n\), then the 10th term of the AP is:
- A. 57
- B. 62
- C. 67
- D. 72
Solution
The nth term of an AP can be found using:
\[
a_n = S_n - S_{n-1}
\]
Given \(S_n = 3n^2 + 5n\):
\[
S_{n-1} = 3(n-1)^2 + 5(n-1) = 3(n^2 - 2n + 1) + 5n - 5 = 3n^2 - 6n + 3 + 5n - 5 = 3n^2 - n - 2
\]
Therefore:
\[
a_n = S_n - S_{n-1} = (3n^2 + 5n) - (3n^2 - n - 2) = 6n + 2
\]
For \(n = 10\):
\[
a_{10} = 6(10) + 2 = 62
\]
So the correct option is B.