Question

The sum of third and seventh term of an AP is 6 and their product is 8. Find the sum of the first 16 terms of the AP.

Answer 1

Let the first term be a and common difference be d
nth term = a+(n-1)d

Given Third Term + Seventh term = (a+2d)+(a+6d) = 6 ==> a+4d = 3
hence, a= 3-4d

Third Term * Seventh term = (a+2d)*(a+6d) = 8
(3-4d+2d)*(3-4d+6d) = 8==> (3-2d)*(3+2d) = 8
i.e. 9-4d^2 = 8==> d^2 = (9-8)/4 = 0.25==> d = 0.5 or -0.5

Now to check which is correct d...
Substitute and find

Case (a): d= 0.5
a+4d = 3==> a=3-4d = 3-4(0.5)=1
3rd term = a+2d= 1+2*0.5 = 2
7th term = a+6d= 1+6*0.5 = 4
Sum = 6 and Product = 8

Case (b): d= -0.5
a+4d = 3==> a=3-4d = 3-4(-0.5) = 3+2 = 5
3rd term = a+2d= 5+2*(-0.5) = 4
7th term = a+6d= 5+6*(-0.5) = 2
Sum = 6 and Product = 8

Since both are matching, we will go with bothvalues

Sum of first 16 terms = n*(2a+(n-1)d)/2 = 16*(2a+15d)/2
= 8*(2a+15d)

Case (a): d= 0.5
Sum = 8*(2*1+15*0.5)=76

Case (b): d= 0.5
Sum = 8*(2*5+15*(-0.5))=20