Question

The sum of first n terms of an A.P. is \(2n^{2} + 13n\). Find its \(n^{th}\) term and hence \(10^{th}\) term.

Hint:

Shortcut for \(S_{n} = An^2 + Bn\):
Common difference \(d = 2A\) and first term \(a = A + B\).
Here \(d = 2(2) = 4\) and \(a = 2 + 13 = 15\).
\(a_{n} = a + (n-1)d = 15 + (n-1)4 = 4n + 11\).

Answer 1

The sum of first n terms of an A.P. is:
\[ S_n = 2n^2 + 13n \] We need to find the nth term \(a_n\) and then the 10th term.

Step 1: Use formula
\[ a_n = S_n - S_{n-1} \]
Compute \(S_{n-1}\):
\[ S_{n-1} = 2(n-1)^2 + 13(n-1) \] Expand:
\[ S_{n-1} = 2(n^2 - 2n + 1) + 13n - 13 \] \[ = 2n^2 - 4n + 2 + 13n - 13 \] \[ = 2n^2 + 9n - 11 \]
Step 2: Find nth term
\[ a_n = S_n - S_{n-1} \] \[ = (2n^2 + 13n) - (2n^2 + 9n - 11) \] \[ = 2n^2 + 13n - 2n^2 - 9n + 11 \] \[ = 4n + 11 \]
So nth term is:
\[ a_n = 4n + 11 \]
Step 3: Find 10th term
\[ a_{10} = 4(10) + 11 = 40 + 11 = 51 \]
Final Answers:
nth term = \(4n + 11\)
10th term = 51