Question 2

We will find the first term and common difference to calculate the sum of the first sixteen terms.

Step 1 — Formulate equations

Let the first term be $a$. Let the common difference be $d$. The third term is $a_3 = a + 2d$. The seventh term is $a_7 = a + 6d$. The sum of the third and seventh terms is 6.

$a_3 + a_7 = 6$

$(a + 2d) + (a + 6d) = 6$

$2a + 8d = 6$

$a + 4d = 3$

$a = 3 - 4d \quad \text{(Equation 1)}$

The product of the third and seventh terms is 8.

$(a_3)(a_7) = 8$

$(a + 2d)(a + 6d) = 8 \quad \text{(Equation 2)}$

Step 2 — Solve for the common difference

We substitute the value of $a$ from Equation 1 into Equation 2.

$((3 - 4d) + 2d)((3 - 4d) + 6d) = 8$

$(3 - 2d)(3 + 2d) = 8$

$3^2 - (2d)^2 = 8$

$9 - 4d^2 = 8$

$4d^2 = 9 - 8$

$4d^2 = 1$

$d^2 = \frac{1}{4}$

$\boxed{d = \pm \frac{1}{2}}$

Step 3 — Find the first term for each case

Case 1: When $d = \frac{1}{2}$

We use Equation 1 to find $a$.

$a = 3 - 4\left(\frac{1}{2}\right)$

$a = 3 - 2$

$\boxed{a = 1}$

Case 2: When $d = -\frac{1}{2}$

We use Equation 1 to find $a$.

$a = 3 - 4\left(-\frac{1}{2}\right)$

$a = 3 + 2$

$\boxed{a = 5}$

Step 4 — Calculate the sum of the first sixteen terms

The formula for the sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$. We need to find $S_{16}$.

Case 1: When $a = 1$ and $d = \frac{1}{2}$

$S_{16} = \frac{16}{2}\left[2(1) + (16-1)\left(\frac{1}{2}\right)\right]$

$S_{16} = 8\left[2 + 15\left(\frac{1}{2}\right)\right]$

$S_{16} = 8\left[2 + \frac{15}{2}\right]$

$S_{16} = 8\left[\frac{4}{2} + \frac{15}{2}\right]$

$S_{16} = 8\left[\frac{19}{2}\right]$

$S_{16} = 4 \times 19$

$\boxed{S_{16} = 76}$

Case 2: When $a = 5$ and $d = -\frac{1}{2}$

$S_{16} = \frac{16}{2}\left[2(5) + (16-1)\left(-\frac{1}{2}\right)\right]$

$S_{16} = 8\left[10 + 15\left(-\frac{1}{2}\right)\right]$

$S_{16} = 8\left[10 - \frac{15}{2}\right]$

$S_{16} = 8\left[\frac{20}{2} - \frac{15}{2}\right]$

$S_{16} = 8\left[\frac{5}{2}\right]$

$S_{16} = 4 \times 5$

$\boxed{S_{16} = 20}$

Answer

The sum of the first sixteen terms of the AP can be 20 or 76.