The sum of the third and the seventh terms of an A.P. is 6 and the product is 8. Find the sum of first sixteen terms of the A.P

Answer

To solve the problem step by step, we need to find the first term (a) and the common difference (d) of the arithmetic progression (A.P.) based on the given conditions about the third and seventh terms.

### Step 1: Write down the formulas for the terms
The nth term of an A.P. can be expressed as:
- \( T_n = a + (n-1)d \)

Thus, we can express the 3rd and 7th terms as:
- \( T_3 = a + 2d \)
- \( T_7 = a + 6d \)

### Step 2: Set up the equations based on the problem statement
According to the problem:
1. The sum of the 3rd and 7th terms is 6:
\[ T_3 + T_7 = 6 \implies (a + 2d) + (a + 6d) = 6 \]
Simplifying this gives:
\[ 2a + 8d = 6 \quad \text{(Equation 1)} \]

2. The product of the 3rd and 7th terms is 8:
\[ T_3 \times T_7 = 8 \implies (a + 2d)(a + 6d) = 8 \]
Expanding this gives:
\[ a^2 + 8ad + 12d^2 = 8 \quad \text{(Equation 2)} \]

### Step 3: Solve Equation 1 for a
From Equation 1:
\[ 2a + 8d = 6 \implies 2a = 6 - 8d \implies a = 3 - 4d \quad \text{(Equation 3)} \]

### Step 4: Substitute Equation 3 into Equation 2
Now substitute \( a \) from Equation 3 into Equation 2:
\[ (3 - 4d)^2 + 8(3 - 4d)d + 12d^2 = 8 \]
Expanding this:
\[ (9 - 24d + 16d^2) + (24d - 32d^2) + 12d^2 = 8 \]
Combining like terms:
\[ 9 + (16d^2 - 32d^2 + 12d^2) = 8 \]
This simplifies to:
\[ 9 - 4d^2 = 8 \]
Rearranging gives:
\[ -4d^2 = -1 \implies 4d^2 = 1 \implies d^2 = \frac{1}{4} \implies d = \frac{1}{2} \text{ or } d = -\frac{1}{2} \]

### Step 5: Find the values of a
Using \( d = \frac{1}{2} \):
\[ a = 3 - 4 \times \frac{1}{2} = 3 - 2 = 1 \]

Using \( d = -\frac{1}{2} \):
\[ a = 3 - 4 \times -\frac{1}{2} = 3 + 2 = 5 \]

### Step 6: Find the sum of the first 16 terms
The sum of the first n terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For \( n = 16 \):
1. Using \( a = 1 \) and \( d = \frac{1}{2} \):
\[ S_{16} = \frac{16}{2} \times (2 \times 1 + (16 - 1) \times \frac{1}{2}) = 8 \times (2 + 7.5) = 8 \times 9.5 = 76 \]

2. Using \( a = 5 \) and \( d = -\frac{1}{2} \):
\[ S_{16} = \frac{16}{2} \times (2 \times 5 + (16 - 1) \times -\frac{1}{2}) = 8 \times (10 - 7.5) = 8 \times 2.5 = 20 \]

### Conclusion
Thus, the sum of the first 16 terms of the A.P. can be either 76 or 20, depending on the values of \( a \) and \( d \).