∵a,b,c, are in G.P. ⇒b2=ac Since, a+b+c=xb ⇒a+c=(x−1)b Take square on both sides, we get a2+c2+2ac=(x−1)2b2 ⇒a2+c2=(x−1)2ac−2ac[∵b2=ac] ⇒a2+c2=ac[(x−1)2−2] ⇒a2+c2=ac[x2−2x−1] ∵a2+c2 are positive and b2=ac which is also positive. Then, x2−2x−1 would be positive but for x=2,x2−2x−1 is negative. Hence, x cannot be taken as 2 .