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

Step 1: Understanding the Concept:
The \(n^{th}\) term of a sequence can be found from the sum of terms using the relation \(a_{n} = S_{n} - S_{n-1}\).
Step 2: Detailed Explanation:
Given: \(S_{n} = 2n^{2} + 13n\)
Find \(S_{n-1}\):
\[ S_{n-1} = 2(n - 1)^{2} + 13(n - 1) \]
\[ S_{n-1} = 2(n^{2} - 2n + 1) + 13n - 13 \]
\[ S_{n-1} = 2n^{2} - 4n + 2 + 13n - 13 = 2n^{2} + 9n - 11 \]
Calculate \(a_{n}\):
\[ a_{n} = S_{n} - S_{n-1} \]
\[ a_{n} = (2n^{2} + 13n) - (2n^{2} + 9n - 11) \]
\[ a_{n} = 2n^{2} + 13n - 2n^{2} - 9n + 11 = 4n + 11 \]
Find \(10^{th}\) term (\(a_{10}\)):
\[ a_{10} = 4(10) + 11 = 40 + 11 = 51 \]
Step 3: Final Answer:
The \(n^{th}\) term is \(4n + 11\) and the \(10^{th}\) term is 51.