The sum of n terms of an A.P. is `3n^(2)+5`. The number of term which equals 159, is

Options

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

Answer

To solve the problem, we need to find the number of terms \( n \) in an arithmetic progression (A.P.) such that the \( n \)-th term equals 159. We are given that the sum of the first \( n \) terms \( S_n \) is given by the formula:

\[ S_n = 3n^2 + 5 \]

### Step 1: Find the expression for the \( n \)-th term \( T_n \)

The \( n \)-th term \( T_n \) of an A.P. can be calculated using the formula:

\[ T_n = S_n - S_{n-1} \]

### Step 2: Calculate \( S_{n-1} \)

To find \( S_{n-1} \), we substitute \( n-1 \) into the sum formula:

\[ S_{n-1} = 3(n-1)^2 + 5 \]

Expanding this:

\[ S_{n-1} = 3(n^2 - 2n + 1) + 5 = 3n^2 - 6n + 3 + 5 = 3n^2 - 6n + 8 \]

### Step 3: Substitute \( S_n \) and \( S_{n-1} \) into the formula for \( T_n \)

Now we can calculate \( T_n \):

\[ T_n = S_n - S_{n-1} = (3n^2 + 5) - (3n^2 - 6n + 8) \]

Simplifying this:

\[ T_n = 3n^2 + 5 - 3n^2 + 6n - 8 \]
\[ T_n = 6n - 3 \]

### Step 4: Set \( T_n \) equal to 159

We need to find \( n \) such that:

\[ 6n - 3 = 159 \]

### Step 5: Solve for \( n \)

Adding 3 to both sides:

\[ 6n = 162 \]

Now, divide both sides by 6:

\[ n = \frac{162}{6} = 27 \]

### Conclusion

The number of terms \( n \) for which the \( n \)-th term equals 159 is:

\[ \boxed{27} \]