In a geometric sequence, the first term is 8 and the sum of the first six terms is 74 648. Determine the third term of the sequence.

A1=8

common ratio=r

sum of 6 terms

S=a1+a2+a3 + ... + a6

=a1 (1+r+r^2 + ... + r^5)

=a1 (r^6-1) / (r-1)

but we're given S=74648

=>

8 (r^6-1) / (r-1) = 74648

Cross multiply and solve for r (by trial and error)

r^6-1=9331 (r-1)

r=6

so

a (3) = a1*r^ (3-1)

=8 * (6^2)

=288