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.

Let first term be a and common difference be d.

By formula,

a_n = a + (n - 1)d

Given,

Sum of the third and the seventh terms of an AP is 6.

⇒ a₃ + a₇ = 6

⇒ a + (3 - 1)d + a + (7 - 1)d = 6

⇒ a + 2d + a + 6d = 6

⇒ 2a + 8d = 6

⇒ 2(a + 4d) = 6

⇒ a + 4d = 3

⇒ a = 3 - 4d ……..(1)

Given,

Product of third and the seventh terms of an AP is 8.

⇒ a₃ × a₇ = 8

⇒ [a + (3 - 1)d][a + (7 - 1)d] = 8

⇒ [a + 2d][a + 6d] = 8

⇒ a² + 6ad + 2ad + 12d² = 8

Substituting value of a from equation (1) in above equation, we get :

⇒ (3 - 4d)² + 6d(3 - 4d) + 2d(3 - 4d) + 12d² = 8

⇒ 9 + 16d² - 24d + 18d - 24d² + 6d - 8d² + 12d² = 8

⇒ 28d² - 32d² - 24d + 24d + 9 - 8 = 0

⇒ -4d² + 1 = 0

⇒ 4d² = 1

⇒ d² = $\dfrac{1}{4}$

⇒ d = $\sqrt{\dfrac{1}{4}}$.

⇒ d = $\pm \dfrac{1}{2}$

Substituting value of $d = \dfrac{1}{2}$ in equation (1), we get :

⇒ a = 3 - 4d

⇒ a = 3 - $4 \times \dfrac{1}{2}$

⇒ a = 3 - 2 = 1.

By formula,

Sum of first n terms = S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$S_n = \dfrac{16}{2}[2 \times 1 + (16 - 1) \times \dfrac{1}{2}] \\[1em] = 8 \times [2 + 15 \times \dfrac{1}{2}] \\[1em] = 8 \times [2 + 7.5] \\[1em] = 8 \times 9.5 \\[1em] = 76.$

Substituting value of $d = -\dfrac{1}{2}$ in equation (1), we get :

⇒ a = 3 - 4d

⇒ a = 3 - $4 \times -\dfrac{1}{2}$

⇒ a = 3 + 2 = 5.

By formula,

Sum of first n terms = S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$S_n = \dfrac{16}{2}[2 \times 5 + (16 - 1) \times -\dfrac{1}{2}] \\[1em] = 8 \times [10 + 15 \times -\dfrac{1}{2}] \\[1em] = 8 \times [10 - 7.5] \\[1em] = 8 \times 2.5 \\[1em] = 20.$

Hence, sum of first sixteen terms of A.P. = 20 or 76.