Question

If a b and c are in GP and
${a}^{ \frac{1}{x} } = {b}^{ \frac{1}{y} } = {c}^{ \frac{1}{z} }$
Prove that x, y, and z are in AP.​

Answer 1

$\fbox{\mathfrak{\large{Answer}}}$

We have,

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

[say]

$= > {a}^{ \frac{1}{x} } = k \\ = > {b}^{ \frac{1}{y} } = k \\ = > {c}^{ \frac{1}{z} } = k$

$= > a = {k}^{x} \\ = > b = {k}^{y} \\ = > c = {k}^{z}$

Now, a, b and c are in GP

$= > {b}^{2} = ac$

$therefore \: ( { {k}^{2}) }^{2} = {k}^{x} . {k}^{y}$

$= > {k}^{2y} = {k}^{x + z}$

$= > 2y = x + z$

Hence x, y, and z are in AP.

Answer 2

SOLUTION

Refer to the attachment.

hope it helps ☺️