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 the 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: Identify the terms in the A.P.
In an A.P., the \( n \)-th term can be expressed as:
\[ a_n = a_1 + (n-1)d \]
where \( a_1 \) is the first term and \( d \) is the common difference.

### Step 2: Write down the given terms.
The terms we need to consider are:
- \( a_1 = a_1 \)
- \( a_4 = a_1 + 3d \)
- \( a_7 = a_1 + 6d \)
- \( a_{10} = a_1 + 9d \)
- \( a_{13} = a_1 + 12d \)
- \( a_{16} = a_1 + 15d \)

### Step 3: Sum these terms.
Now we can sum these terms:
\[ a_1 + (a_1 + 3d) + (a_1 + 6d) + (a_1 + 9d) + (a_1 + 12d) + (a_1 + 15d) \]
This simplifies to:
\[ 6a_1 + (3 + 6 + 9 + 12 + 15)d = 6a_1 + 45d \]

### Step 4: Set the sum equal to 147.
According to the problem, this sum equals 147:
\[ 6a_1 + 45d = 147 \]

### Step 5: Simplify the equation.
We can divide the entire equation by 3 to simplify:
\[ 2a_1 + 15d = 49 \]

### Step 6: Find the sum of the first 16 terms.
The sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a_1 + (n-1)d) \]
For \( n = 16 \):
\[ S_{16} = \frac{16}{2} \times (2a_1 + 15d) = 8 \times (2a_1 + 15d) \]

### Step 7: Substitute the value from Step 5.
From Step 5, we know that \( 2a_1 + 15d = 49 \):
\[ S_{16} = 8 \times 49 = 392 \]

### Conclusion:
The sum of the first 16 terms of the A.P. is:
\[ \boxed{392} \]