Question

If $2(y-a)$ is the H.M between $y-x$ and $y-z$, then $x-a,y-a,z-a$ are in

• A. A.P
• B. G.P
• C. H.P
• D. None of these

Answer 1

The correct option is B

G.P

Explanation for the correct option:

Given that, $(y–x),2(y–a),(y–z)$ are in HP.

Now, $\frac{1}{(y-x)},\frac{1}{2(y-a)},\frac{1}{(y-z)}$ are in AP.

$\Rightarrow$$\frac{1}{2(y-a)}–\frac{1}{(y-x)}=\frac{1}{(y-z)}–\frac{1}{2(y-a)}$

$\Rightarrow$ $\frac{2a–y–x}{y–x}=\frac{y+z–2a}{y–z}$

$\Rightarrow$ $\frac{(x-a)+(y-a)}{(x-a)-(y-a)}=\frac{(y-a)+(z-a)}{(y-a)-(z-a)}$

$\Rightarrow$ $\frac{x–a}{y–a}=\frac{y–a}{z–a}$

$\Rightarrow$ ${(y–a)}^2=(x–a)(z–a)$

$\therefore$$(x–a),(y–a),(z–a)$ are in GP.

Hence, Option ‘B’ is Correct.