Question

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.

Answer 1

Let the 3^rd term of AP = a + 2d 

and the 7^th term of AP = a + 6d 

∴ Sum of 3^rd and 7^th terms = a + 2d + a + 6d = 6 

⇒ 2a + 8d = 6 

⇒ a + 4d = 3 …… (1) 

Now product of above two terms = (a + 2d) (a + 6d) = 8 

we can re-write above terms as following 

(a + 4d – 2d) (a + 4d + 2d) = 8 

⇒ (3 – 2d) (3 + 2d) = 8 

⇒ 9 – 4d² = 8 

⇒ 4d² = 9 – 8 = 1 

∴ d² = \(\frac{1}{4}\)

⇒ d = ± \(\frac{1}{2}\)……. (2) 

Now putting d = \(\frac{1}{2}\) in eq(1) we get 

a + 4d = a + 4(\(\frac{1}{2}\)) = 3 ⇒ a = 1

so a = 1, d = \(\frac{1}{2}\)

(or) now putting d = \(\frac{-1}{2}\) , we get 

a + 4d = a + 4(\(\frac{-1}{2}\)) = 3 

⇒ a – 2 = 3 

⇒ a = 5 

∴ a = 5, d = \(\frac{-1}{2}\)

∴ Sum of sixteen terms =

So S₁₆ = 20 or 76