Question

If a, b, c are three distinct real numbers in G.P. and a + b + c = xb, then prove that either x < −1 or x > 3.

Answer 1

$$\begin{gathered} \mathrm{Let}r\mathrm{be}\mathrm{the}\mathrm{common}\mathrm{ratio}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{G}.\mathrm{P}. \\ \therefore b=ar\mathrm{and}c=a{r}^2 \\ \mathrm{Now},a+b+c=bx \\ \Rightarrow a+ar+a{r}^2=arx \\ \Rightarrow r^2+\left(1-x\right)r+1=0 \\ r\mathrm{is}\mathrm{always}\mathrm{a}\mathrm{real}\mathrm{number}. \\ \therefore \mathrm{D}\ge 0 \\ \Rightarrow {\left(1-x\right)}^2-4\ge 0 \\ \Rightarrow x^2-2x-3\ge 0 \\ \Rightarrow \left(x-3\right)\left(x+1\right)\ge 0 \\ \Rightarrow x>3\mathrm{or}x<-1\mathrm{and}x\ne 3\mathrm{or}-1\left[\because \mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{are}\mathrm{distinct}\mathrm{real}\mathrm{numbers}\right] \end{gathered}$$