Question

The sum of 3rd and 7th term of an ap is 6 and there product is 8. find the sum of first 16 terms of the ap

Answer 1

Let a and d be the first term and common difference of A.P.

nth term of A.P., an = a + (n – 1) d

∴  a3 = a + (3 – 1) d = a + 2d

a7 = a + (7 – 1) d = a + 6d

Given, a3 + a7 = 6

∴ (a + 2d) + (a + 6d) = 6

⇒ 2a + 8d = 6

⇒ a + 4d = 3     ...(1)

Given, a3 × a7 = 8

∴ (a + 2d) + (a + 6d) = 8

⇒ (3 – 4d + 2d) (3 – 4d + 6d) = 8             [ Using (1) ]

⇒ (3 – 2d) (3 + 2d) = 8

⇒ 9 – 4d2 = 8

⇒ 4d2 = 1

a = 3 – 4d

= 3 – 2 = 1

When a = 1 and

Thus, the sum of first 16 terms of the A.P. is 76 or 20.

Answer 2

Here is your solution

Given:-

The sum of 3rd term and 7th term
=>a+2d+a+6d = 6
=>2a+8d = 6
After common taking
a + 4d = 3
a=3-4d ..................….......(1)

Their product is

(a+2d)×(a+6d) = 8 .........…(2)

Putting value of a = 3–4d in equation (2) to get

=>(a+2d)×(a+6d) = 8

=>(3–4d+2d)×(3–4d+6d) =8

=>(3–2d)×(3+2d) = 8

=>9–4d^2 = 8

=>4d^2 = 9-8

=>4d^2 = 1

=>d^2 = 1/4

=>d = 1/2.

From (1) a = 3–4d
a=> 3–4×1/2
a=>3-2
a=> 1

The sum of the first 16 terms

Sum of 16 terms = (n/2)[2a+(n-1)d]

= (16/2)[2×1 + (16–1)×1/2]

= 8[2+15/2]

= 8×19/2

= 76

Hence,
The sum of first 16 term is 76.

Hope it helps you