Question

If ab + c²; a $-$ b + c = b + c ; 1, then prove that b is the geometric mean between a and c.

Answer 1

$$\begin{gathered} \frac{ab+c^2}{a-b+c}=\frac{b+c}{1}\left(\mathrm{Given}\right) \\ \mathrm{By}\mathrm{cross}-\mathrm{multiplying},\mathrm{we}\mathrm{get}: \\ ab+c^2=\left(b+c\right)\left(a-b+c\right) \\ \Rightarrow ab+c^2=ab-b^2+bc+ac-bc+c^2 \\ \Rightarrow b^2=ac \\ \mathrm{Therefore},b\mathrm{is}\mathrm{the}\mathrm{geometric}\mathrm{mean}\mathrm{between}a\mathrm{and}c. \end{gathered}$$