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