Q2:

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

Given: An Arithmetic Progression (AP) where the sum of the third term ($a_3$) and the seventh term ($a_7$) is $6$, and their product is $8$.

To Find: The sum of the first sixteen terms ($S_{16}$) of the AP.

Step 1: Defining the variables and formulas
Let the first term of the AP be $a$ and the common difference be $d$.
The $n^{th}$ term of an AP is given by the formula: $a_n = a + (n - 1)d$.
The sum of the first $n$ terms of an AP is given by: $S_n = \frac{n}{2} [2a + (n - 1)d]$.

Step 2: Expressing the given conditions algebraically
Using the formula for the $n^{th}$ term:
$a_3 = a + (3 - 1)d = a + 2d$
$a_7 = a + (7 - 1)d = a + 6d$

Condition 1: Sum of the terms is $6$
$(a + 2d) + (a + 6d) = 6$
$2a + 8d = 6$
Dividing by $2$: $a + 4d = 3 \implies a = 3 - 4d$ --- (Equation 1)

Condition 2: Product of the terms is $8$
$(a + 2d)(a + 6d) = 8$ --- (Equation 2)

Step 3: Solving for $a$ and $d$
Substitute Equation 1 into Equation 2:
$((3 - 4d) + 2d)((3 - 4d) + 6d) = 8$
$(3 - 2d)(3 + 2d) = 8$
Using the algebraic identity $(x - y)(x + y) = x^2 - y^2$:
$3^2 - (2d)^2 = 8$
$9 - 4d^2 = 8$
$-4d^2 = 8 - 9$
$-4d^2 = -1$
$d^2 = \frac{1}{4} \implies d = \pm \frac{1}{2}$

Case 1: If $d = \frac{1}{2}$
$a = 3 - 4(\frac{1}{2}) = 3 - 2 = 1$

Case 2: If $d = -\frac{1}{2}$
$a = 3 - 4(-\frac{1}{2}) = 3 + 2 = 5$

Step 4: Calculating $S_{16}$ for both cases
The formula for $S_{16}$ is $S_{16} = \frac{16}{2} [2a + (16 - 1)d] = 8[2a + 15d]$.

For Case 1 ($a=1, d=1/2$):
$S_{16} = 8[2(1) + 15(1/2)] = 8[2 + 7.5] = 8[9.5] = 76$

For Case 2 ($a=5, d=-1/2$):
$S_{16} = 8[2(5) + 15(-1/2)] = 8[10 - 7.5] = 8[2.5] = 20$

Final Answer: The sum of the first sixteen terms is either 76 or 20.