The sum of the first three terms of a geometric sequence is equal to 42

The sum of the first three terms of a geometric sequence is equal to 42

The sum of the first three terms of a geometric sequence is equal to 42. The sum of the squares of the same terms is equal to 1092. Find the three terms of the sequence.

Answer：
sum1 = a + a r + a r2 = 42: the sum of the three terms given, r is the common ratio.
sum2 = a2 + a2r2 + ar2r4 = 1092: the sum of the squares of the three terms given .
sum1 = a + ar + ar2 = a(r3 - 1) / (r - 1) = 42 : apply formula for a finite sum of geometric series.
sum2 = a2 + a2r2 + ar2r4 = a2(r6 - 1) / (r2 - 1) = 1092: the sum of squares is a also a sum of geometric series.
sum2/sum12 = 1092 / 422 = [ a2(r6 - 1)/(r2 - 1)] / [a2(r3 - 1)2 / (r - 1)2]
(r2 - r + 1) / (r2 + r + 1) = 1092 / 422
r = 4 , r = 1/4 : solve for r
a = 2 : substitute r = 4 and solve for a
a = 32 : substitute r = 1/4 and solve for a
a = 2 , ar = 8 , ar2 = 32 : find the three terms for r = 4
a = 32 , ar = 8 , ar2 = 2 : find the three terms for r = 1/4