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 step by step, we need to find the term in the arithmetic progression (A.P.) that equals 159, given that the sum of the first n terms \( S_n \) is \( 3n^2 + 5 \).

### Step 1: Use the formula for the nth term of an A.P.
The nth term \( T_n \) of an A.P. can be calculated using the formula:
\[ T_n = S_n - S_{n-1} \]
where \( S_n \) is the sum of the first n terms and \( S_{n-1} \) is the sum of the first \( n-1 \) terms.

### Step 2: Calculate \( S_{n-1} \)
Given \( S_n = 3n^2 + 5 \), we need to find \( S_{n-1} \):
\[ S_{n-1} = 3(n-1)^2 + 5 \]
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 substitute \( S_n \) and \( S_{n-1} \) into the formula for \( T_n \):
\[ T_n = S_n - S_{n-1} = (3n^2 + 5) - (3n^2 - 6n + 8) \]
Simplifying this:
\[ T_n = 3n^2 + 5 - 3n^2 + 6n - 8 = 6n - 3 \]

### Step 4: Set \( T_n \) equal to 159
Now we need to find \( n \) such that:
\[ T_n = 159 \]
So we set up the equation:
\[ 6n - 3 = 159 \]

### 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 term which equals 159 is the 27th term of the A.P.

### Final Answer
The number of terms which equals 159 is **27**.

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