Question

The sum of 3rd and 7th term of an AP is 6 and their product is 8. Find the sum of 1st sixteen terms of the AP.

Answer 1

As it is an AP, let first term be 'a' and common difference be 'd'.
Therefore, 3rd term =a+(3-1)d =a+2d
Similarly, 7th term =a+6d
But it is given that their sum is 6, therefore–
(a+2d)+(a+6d)=6
=>2a+8d=6
=>a+4d=3
On squaring;
=>a^2+8ad+16d^2=9
=>a^2+8ad+12d^2=9-4d^2 (why we did this?)
It is also given that their product is 8, so–
(a+2d)(a+6d)=8
=>a^2+12d^2+8ad=8
Or, from above,
=>9-4d^2=8
=>4d^2=1
=>d^2=1/4
=>d=1/2 or -1/2
Taking d=1/2
a+4(1/2)=3
=>a+2=3
=>a=3-2=1
We can also verify that 3rd and 7th term multiply to 8.
[1+2×(1/2)][1+6×(1/2)] =2×4 =8
Hence, 1st term is 1 and common difference is 1/2.
Sum of n terms in AP is (n/2)[2a+(n-1)d] =(16/2)[2×1+(16-1)×(1/2)]
=8×[2+15/2]
=16+60
=76 [Answer]
Hope this makes some sense!

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.