Question

If a, b, c are in A.P., b,c,d are in G.P. and $\frac{1}{c},\frac{1}{d},\frac{1}{e}$ are in A.P., prove that a, c,e are in G.P.

Answer 1

$$\begin{gathered} a,b\mathrm{and}c\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \therefore 2b=a+c.......\left(\mathrm{i}\right) \\ \mathrm{Also},b,c\mathrm{and}d\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}. \\ \therefore c^2=bd.......\left(\mathrm{ii}\right) \\ \mathrm{And}\frac{1}{c},\frac{1}{d}\mathrm{and}\frac{1}{e}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \therefore \frac{2}{d}=\frac{1}{c}+\frac{1}{e} \\ \Rightarrow d=\frac{2ce}{c+e}.......\left(\mathrm{iii}\right) \\ \because c^2=bd\left[\mathrm{From}\left(\mathrm{ii}\right)\right] \\ \Rightarrow c^2=\left(\frac{a+c}{2}\right)\left(\frac{2ce}{c+e}\right)\left[\mathrm{Using}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{iii}\right)\right] \\ \Rightarrow c^2\left(c+e\right)=ce\left(a+c\right) \\ \Rightarrow c^2+ce=ae+ec \\ \Rightarrow c^2=ae \\ \mathrm{Therefore},a,c\mathrm{and}e\mathrm{are}\mathrm{also}\mathrm{in}\mathrm{G}.\mathrm{P}. \end{gathered}$$