If $a^{\frac {1}{x}}= b^{\frac {1}{y}}=c^{\frac {1}{z}}$ and $a$, $b$, $c$ are in AP then $x$, $y$, $z$ are in

Question

If $a^{\frac {1}{x}}= b^{\frac {1}{y}}=c^{\frac {1}{z}}$ and $a$, $b$, $c$ are in AP then $x$, $y$, $z$ are in

$a$. A.P

$b$. G.P

$c$. H.P

$d$. None.

My Attempt: $a$, $b$, $c$ are in AP $$\dfrac {a+c}{2}=b$$ $$a+c=2b.$$ Now, $$a^{\dfrac {1}{x}}=b^{\dfrac {1}{y}}$$ $$a^{\dfrac {y}{x}}=b$$ $$2a^{\dfrac {y}{x}}=2b$$ $$2a^{\dfrac {y}{x}}=a+c$$

How do I proceed?

Answer 1

There is a typo in your question. It should be "If $a, b, c$ are in G.P. and $a^{1 / x}=b^{1 / y}=c^{1 / t}$, then show that $x, y, z$ are in A.P."

Let $a^{1 / x}=b^{1 / y}=c^{1 / 4}=k$. Then, $a=k^{x}, b=h^{y}$ and $c=k^{z}$. Now, $a, b, c$ are in G.P. $\Rightarrow b^{2}=a c$ $$ \begin{aligned} \Rightarrow\left(k^{y}\right)^{2}=k^{x}, h^{2}=h^{x+2} \\ & \Rightarrow k^{2 y}=k^{x+z} \\ & \Rightarrow 2 y=(x+z) . \end{aligned} $$ This shows that $x, y, z$ are in A.P.