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, the first term = a and common difference = d
the sum of the third and the seventh terms of the AP is 6, therefore
a₃ + a₇ = 6
⇒ a + 2d + a + 6d = 6
⇒ 2a + 8d = 6
⇒ a + 4d = 3
⇒ a = 3 – 4d …(1)
The product of the third and the seventh terms of the AP is 8, therefore
(a₃)(a₇) = 8
⇒ (a + 2d)(a + 6d) = 8
Putting the value of a from the equation(1), we get
(3 -2d)(3 +2d) = 8
⇒ 3² – 4d² = 8
⇒ 4d² = 1
⇒ d = ± 1/2
if, d = 1/2,
Putting the value of d in the equation (1), we get
a = 3 – 4(1/2) = 1
The sum of the 16 terms of this AP is given by
S₁₆ = 16/2[2a + 16 – 1)d] = 8[2(1) + 15 (1/2)] = 76
If, d = – 1/2,
Putting the value of d in the equation (1), we get
a = 3 – 4(-1/2) = 5
The sum of the 16 terms of this AP is given by
S₁₆ = 16/2[2a + 16 – 1)d] = 8 [2(5) + 15 (-1/2)] = 20
Hence, the sum of the 16 terms of AP is 20 or 76.