Find the sum of first 16 terms of an A.P. `a _(1), a _(2), a_(3)………..`
If it is known that `a _(1) + a _(4) + a _(7) + a _(10) + a _(13) + a _(16) = 147`

Answer

To find the sum of the first 16 terms of an arithmetic progression (A.P.) given that \( a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = 147 \), we can follow these steps:

### Step 1: Define the terms of the A.P.
The \( n \)-th term of an A.P. can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Write the specific terms involved in the sum
We need to express \( a_1, a_4, a_7, a_{10}, a_{13}, a_{16} \):
- \( a_1 = a \)
- \( a_4 = a + 3d \)
- \( a_7 = a + 6d \)
- \( a_{10} = a + 9d \)
- \( a_{13} = a + 12d \)
- \( a_{16} = a + 15d \)

### Step 3: Set up the equation based on the given information
Now we can sum these terms:
\[ a_1 + a_4 + a_7 + a_{10} + a_{13} + a_{16} = a + (a + 3d) + (a + 6d) + (a + 9d) + (a + 12d) + (a + 15d) \]
This simplifies to:
\[ 6a + (3 + 6 + 9 + 12 + 15)d = 147 \]
Calculating the sum of the coefficients of \( d \):
\[ 3 + 6 + 9 + 12 + 15 = 45 \]
So we have:
\[ 6a + 45d = 147 \]

### Step 4: Simplify the equation
Dividing the entire equation by 3 gives:
\[ 2a + 15d = 49 \quad \text{(Equation 1)} \]

### Step 5: Find the sum of the first 16 terms
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For \( n = 16 \):
\[ S_{16} = \frac{16}{2} \times (2a + 15d) = 8 \times (2a + 15d) \]
Substituting \( 2a + 15d \) from Equation 1:
\[ S_{16} = 8 \times 49 = 392 \]

### Final Answer
Thus, the sum of the first 16 terms of the A.P. is:
\[ \boxed{392} \]