If a,b,c are three distinct real numbers in G.P.

Question

If $a,b,c$ are three distinct real numbers in G.P. and $a+b+c=xb$, then $x$ cannot be:

Accepted Answer

$b^2=ac$; $a+c=(x-1)b$. Squaring: $a^2+c^2+2ac=(x-1)^2b^2 \Rightarrow a^2+c^2 = ac\big[(x-1)^2-2\big] = ac(x^2-2x-1).$

Since $a^2+c^2>0$ and $ac=b^2>0$, we need $x^2-2x-1>0$. At $x=2$: $4-4-1=-1<0$, so $x=2$ is impossible.

Solutions · 1