Let the first term be $a$ and common difference be $d$.
Setting up equations:
$a_3 = a + 2d$ and $a_7 = a + 6d$
Given: $a_3 + a_7 = 6$
$$( a + 2d) + (a + 6d) = 6 \implies 2a + 8d = 6 \implies a + 4d = 3 \quad \text{...(1)}$$
Given: $a_3 \times a_7 = 8$
$$(a + 2d)(a + 6d) = 8 \quad \text{...(2)}$$
Solving:
From (1): $a = 3 - 4d$. Substituting in (2):
$$(3 - 4d + 2d)(3 - 4d + 6d) = 8$$
$$(3 - 2d)(3 + 2d) = 8$$
$$9 - 4d^2 = 8 \implies 4d^2 = 1 \implies d = \pm\frac{1}{2}$$
Case 1: $d = \frac{1}{2}$, then $a = 3 - 4\times\frac{1}{2} = 1$
Case 2: $d = -\frac{1}{2}$, then $a = 3 - 4\times\left(-\frac{1}{2}\right) = 5$
Sum of first 16 terms:
$$S_{16} = \frac{16}{2}[2a + 15d] = 8[2a + 15d]$$
Case 1: $S_{16} = 8\left[2(1) + 15\times\frac{1}{2}\right] = 8\left[2 + 7.5\right] = 8 \times 9.5 = \mathbf{76}$
Case 2: $S_{16} = 8\left[2(5) + 15\times\left(-\frac{1}{2}\right)\right] = 8\left[10 - 7.5\right] = 8 \times 2.5 = \mathbf{20}$
∴ The sum of the first 16 terms is 76 or 20.
Source: Chapter 5, Exercise 5.4 Q.2
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