Let ‘a’ be the first term & ‘x’ be the common ratio.Also, suppose 27, 8 & 12 be the pth, qth & rth term of the GP. ∴axp−1=27 axq−1=8 and axr−1=12 Now,27×82=123 ⇒axp−1x(axq−1)2=(axr−1)3 ⇒xp−1⋅x2q−2=x2r−3 ⇒p−1+2q−2=3r−3 ⇒p+2q−3r=0...(1) There are infinitely many solutions for the eq. (1).