Sry...i don't know...qki m abhi 3rd... chapter he bana rhi hu....ap....ka a bhi nhi janti hu....!!!we r not padhne wala baccha...n....bs glti se acche number aa jate h....

Given Information
The problem states that the sum of the 3rd and 7th terms of an arithmetic progression (AP) is 6, and their product is 8. We need to find the sum of the first sixteen terms of this AP.
Defining the Terms
- The nth term of an AP can be expressed as:
\[ T_n = a + (n-1)d \]
where 'a' is the first term and 'd' is the common difference.
- Therefore, the 3rd and 7th terms are:
\[ T_3 = a + 2d \]
\[ T_7 = a + 6d \]
Setting Up the Equations
- From the given information, we can formulate two equations:
1. Sum of terms:
\[ T_3 + T_7 = (a + 2d) + (a + 6d) = 2a + 8d = 6 \]
2. Product of terms:
\[ T_3 \cdot T_7 = (a + 2d)(a + 6d) = 8 \]
Simplifying the Equations
- From the sum equation, we can rearrange it:
\[ 2a + 8d = 6 \]
\[ a + 4d = 3 \]
\[ a = 3 - 4d \]
- Substituting 'a' into the product equation:
\[ (3 - 4d + 2d)(3 - 4d + 6d) = 8 \]
Simplifying gives:
\[ (3 - 2d)(3 + 2d) = 8 \]
- This expands to:
\[ 9 - 4d^2 = 8 \]
\[ 4d^2 = 1 \]
\[ d^2 = \frac{1}{4} \]
\[ d = \frac{1}{2} \text{ or } -\frac{1}{2} \]
Finding 'a' and the Sum of the First 16 Terms
- If \( d = \frac{1}{2} \):
\[ a = 3 - 4(\frac{1}{2}) = 1 \]
- If \( d = -\frac{1}{2} \):
\[ a = 3 - 4(-\frac{1}{2}) = 5 \]
- The sum of the first 'n' terms of an AP is given by:
\[ S_n = \frac{n}{2} [2a + (n - 1)d] \]
- For \( n = 16 \):
1. If \( a = 1 \) and \( d = \frac{1}{2} \):
\[ S_{16} = \frac{16}{2} [2(1) + 15(\frac{1}{2})] = 8 [2 + 7.5] = 8 \