Question

hey

100 points

Plz solve this with clear explanation

☆ Best answer will be marked as brainliest ​

Answer 1

Here is your ans..

Hope it helps....

Mark as brainliest.....

Answer 2

Hey mate!

Here's your answer!

It is given that,

a3+a7 = 6

=> a+2d+a+6d = 6

=> 2a+8d = 6

=> a+4d = 3

=> a = 3-4d ...(i)

Also,

a3×a7 = 8

=> (a+2d)×(a+6d) = 8 ...(ii)

Substituting (i) in (ii),

(3-4d+2d)×(3-4d+6d) = 8

=> (3-2d)×(3+2d) = 8

$= > 9 - 4d {}^{2} = 8 \\ = > d {}^{2} = 1 \div 4 \\ = > d = 1 \div 2$

When d = 1/2,

a = 3-4(1/2)

=> a = 3-2

=> a = 1

We know that,

Sn = n/2[2a+(n-1)d]

So,

S16 = 16/2[2(1)+15(1/2)]

=> 8[(4+15)/2]

=> 8(19/2)

=> 4(19)

=> 76.

Hence, the sum of the first sixteen terms of the AP is 76.

Hope it helps :)