Question

If a, b, c are in G.P. and x, y are AM's between a, b and b,c respectively, then

(a) $\frac{1}{x}+\frac{1}{y}=2$

(b) $\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$

(c) $\frac{1}{x}+\frac{1}{y}=\frac{2}{a}$

(d) $\frac{1}{x}+\frac{1}{y}=\frac{2}{b}$

Answer 1

(d) $\frac{1}{x}+\frac{1}{y}=\frac{2}{b}$

$$\begin{gathered} a,b\mathrm{and}c\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}. \\ \therefore b^2=ac........\left(\mathrm{i}\right) \\ a,x\mathrm{and}b\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \therefore 2x=a+b........\left(\mathrm{ii}\right) \\ \mathrm{Also},b,y\mathrm{and}c\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \therefore 2y=b+c \\ \Rightarrow 2y=b+\frac{b^2}{a}\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ \Rightarrow 2y=b+\frac{b^2}{\left(2x-b\right)}\left[\mathrm{Using}\left(\mathrm{ii}\right)\right] \\ \Rightarrow 2y=\frac{b\left(2x-b\right)+b^2}{\left(2x-b\right)} \\ \Rightarrow 2y=\frac{2bx-b^2+b^2}{\left(2x-b\right)} \\ \Rightarrow 2y=\frac{2bx}{\left(2x-b\right)} \\ \Rightarrow y=\frac{bx}{\left(2x-b\right)} \\ \Rightarrow y\left(2x-b\right)=bx \\ \Rightarrow 2xy-by=bx \\ \Rightarrow bx+by=2xy \\ \mathrm{Dividing}\mathrm{both}\mathrm{the}\mathrm{sides}\mathrm{by}xy: \\ \Rightarrow \frac{1}{y}+\frac{1}{x}=\frac{2}{b} \end{gathered}$$