Question

the sum of the 3rd and 7th term of an a.p. is 6 and their product is 8 . find the sum of first sixteenterms of the a.p.

Answer 1

Here is your answer
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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.

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