If a, b, c are in G.P. and a

If abc are in G.P. and a1/b1/y = c1/z, then xyz are in

(a) AP

(b) GP

(c) HP

(d) none of these


(a) AP

$a, b$ and $c$ are in G.P.

$\therefore b^{2}=a c$

Taking log on both the sides :

$2 \log \mathrm{b}=\log \mathrm{a}+\log \mathrm{c}$          ….(i)

Now, $a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}$

Taking log on both the side $s:$

$\frac{\log \mathrm{a}}{\mathrm{x}}=\frac{\log \mathrm{b}}{\mathrm{y}}=\frac{\log \mathrm{c}}{\mathrm{z}} \quad \ldots \ldots$ (ii)

Now, comparing (i) and (ii):

$\frac{\log a}{x}=\frac{\log \mathrm{a}+\log \mathrm{c}}{2 y}=\frac{\log \mathrm{c}}{\mathrm{z}}$

$\Rightarrow \frac{\log a}{x}=\frac{\log \mathrm{a}+\log \mathrm{c}}{2 y}$ and $\frac{\log \mathrm{a}}{\mathrm{x}}=\frac{\log \mathrm{c}}{\mathrm{z}}$

$\Rightarrow \log \mathrm{a}(2 \mathrm{y}-\mathrm{x})=\mathrm{x} \log \mathrm{c}$ and $\frac{\log \mathrm{a}}{\log \mathrm{c}}=\frac{\mathrm{x}}{\mathrm{z}}$

$\Rightarrow \frac{\log \mathrm{a}}{\log \mathrm{c}}=\frac{\mathrm{x}}{(2 \mathrm{y}-\mathrm{x})}$ and $\frac{\log \mathrm{a}}{\log \mathrm{c}}=\frac{\mathrm{x}}{\mathrm{z}}$

$\Rightarrow \frac{\mathrm{x}}{(2 \mathrm{y}-\mathrm{x})}=\frac{\mathrm{x}}{\mathrm{z}}$

$\Rightarrow 2 y=x+z$

Thus, $x, y$ and $z$ are in A.P.






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