Question

Let x be the A.M. and y, z be two G.M.s between two positive numbers. Then, $\frac{y^3+z^3}{xyz}$ is equal
to
(a) 1
(b) 2
(c) $\frac{1}{2}$
(d) none of these

Answer 1

(b) 2

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{two}\mathrm{numbers}\mathrm{be}a\mathrm{and}b. \\ a,x\mathrm{and}b\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \therefore 2x=a+b\left(\mathrm{i}\right) \\ \mathrm{Also},a,y,z\mathrm{and}b\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}. \\ \therefore \frac{y}{a}=\frac{z}{y}=\frac{b}{z} \\ \Rightarrow y^2=az,yz=ab,z^2=by\left(\mathrm{ii}\right) \\ \mathrm{Now},\frac{y^3+z^3}{xyz} \\ =\frac{y^2}{xz}+\frac{z^2}{xy} \\ =\frac{1}{x}\left(\frac{y^2}{z}+\frac{z^2}{y}\right) \\ =\frac{1}{x}\left(\frac{az}{z}+\frac{by}{y}\right)\left[\mathrm{Using}\left(\mathrm{ii}\right)\right] \\ =\frac{1}{x}\left(a+b\right) \\ =\frac{2}{\left(a+b\right)}\left(a+b\right)\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ =2 \end{gathered}$$