The sum of the first n terms sn = a1* (1-q n)/(1-q) = 48.

but

s2n=a 1*( 1-q^(2n))/( 1-q)=60

Division of two formulas

1/( 1+q^n)=48/60

Then q n = 1/4.

The sum of the first 3n terms is

s3n=a 1*( 1-q^3n)/( 1-q)=a 1*( 1-q^n)( 1+q^n+q^2n)/( 1-q)

=S2n*( 1+q^n+q^2n)

=48*( 1+ 1/4+ 1/ 16)

=63

In fact, the first n sums, the last n sums and the last n sums are also geometric series.

Sum every n terms, and the number formed is also a geometric series.

That is:

(S2n-Sn)^2=Sn*(S3n-S2n)