Question

The sum of 3rd n 7th term of an AP is 6 and their product is 8. Find the sum of first 16 terms of the AP

Answer 1

given:a3+a7=6 and a3*a7=8
Now,
a3=a+2d and a7=a+6d
=>a3+a7=a+2d+a+6d
=2a+8d
=>2a+8d=6
=>a+4d=3.......(1)

a3*a7=(a+2d)(a+6d)
=a^2+8ad+12d^2
=>a^2+8ad+12d^2=8.......(2)

sqauring (1) we get:
a^2+16d^2+8ad=9......(3)

now (3)-(2) gives:
4d^2=1
=>d=1/2

now we know that a+4(d)=3
so from this equation a=1

now Sn=(n/2){2a+(n-1)d}
so S16=(16/2){2*1+(15)*(1/2)}
solving it we get S16=76