Question

Sum and product of the third and seventh term of an AP are 6 and 8 respectively find the sum of the first 16 term of AP.

Answer 1

Answer:

a3+a7=a+2d+a+6d are sum

a3*a7=(a+2d)(a+6d)

sum=a+2d+a+6d=6

=2a+8d=6

divide whole by 2,we get

a+4d=3.........(1)

a=3-4d eq(1)...

now solving product of (a+2d)(a+6d)=8

putting eq (1) on eq(2)

we get (3-4d+2d)(3-4d +6d)=8

=(3-2d) (3+2d)=8

=3(3+2d)-2d(3+2d)=8

9+6d-6d-4d^2=8

=9-4d^2=8

=-4d^2=8-9

=-4d^2=-1

=d^2=1/4

hence d=1/2,

a+2d+a+6d =6

2a+8(1/2)=6

2a=2

a=1

sum of 16 term of ap=Sn=n/2(2a+(n-1)d)

S16=16/2(2(1)+(16-1)1/2

S16=8(2+15/2)

S16=8(19/2)

S16=76