Using summation formula of G.P. having 1st term as a and common ratio as r we have,
a(r3–1)(r–1)=42 ……..(A)
Squaring the terms of the G.P. yields another G.P. having 1st term as a2 and common ratio as r2. Thus,
a2(r6–1)(r2–1)=1092
⇒a2(r3–1)(r3+1)(r–1)(r+1)=1092
Using (A), 42×a(r3+1)(r+1)=1092
⇒a(r2–r+1)=26
Given, a(r2+r+1)=42
∴r2+r+1r2–r+1=4226=2113
⇒8r2–34r+8=0 or 4r2–17r+4=0
⇒r=4,1/4
Using, a=26r2–r+1;a=2,32
The 3 terms of G.P. are either 2, 8, 32 or 32, 8, 2