The sum of n terms of an A.P. is `3n^(2) + 5`. If `T_(n)` of the series is 159, then n equals

Options

A. 13
B. 21
C. 27
D. none of these

Answer

To solve the problem, we need to find the value of \( n \) given that the sum of the first \( n \) terms of an arithmetic progression (A.P.) is \( S_n = 3n^2 + 5 \) and the \( n \)-th term \( T_n \) is 159.

### Step-by-Step Solution:

1. **Identify the formula for the \( n \)-th term**:
The \( n \)-th term \( T_n \) of an A.P. can be expressed in terms of the sum of the first \( n \) terms:
\[ T_n = S_n - S_{n-1} \]
where \( S_{n-1} \) is the sum of the first \( n-1 \) terms.

2. **Calculate \( S_{n-1} \)**:
We know that:
\[ S_n = 3n^2 + 5 \]
To find \( S_{n-1} \), we substitute \( n-1 \) into the sum formula:
\[ S_{n-1} = 3(n-1)^2 + 5 = 3(n^2 - 2n + 1) + 5 = 3n^2 - 6n + 3 + 5 = 3n^2 - 6n + 8 \]

3. **Substitute into the formula for \( T_n \)**:
Now we can express \( T_n \):
\[ T_n = S_n - S_{n-1} = (3n^2 + 5) - (3n^2 - 6n + 8) \]
Simplifying this gives:
\[ T_n = 3n^2 + 5 - 3n^2 + 6n - 8 = 6n - 3 \]

4. **Set \( T_n \) equal to 159**:
We know that \( T_n = 159 \):
\[ 6n - 3 = 159 \]

5. **Solve for \( n \)**:
Add 3 to both sides:
\[ 6n = 162 \]
Now, divide by 6:
\[ n = \frac{162}{6} = 27 \]

### Final Answer:
The value of \( n \) is \( 27 \).