If x - 1- y-1- z-1 are in GP- then 1-1-ln x- 1-1-ln y- 1-1-ln z are in

Let the common ratio of the GP be r. Then, y=xr and z =xr^2 ⇒ ln y = ln x+ln r and ln z = ln x+2 ln r Let A = 1 + ln x, D = ln r Then, 1/1+ln x = 1/A, 1/1+ln ...

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Byju's Answer

Standard XII

Mathematics

Harmonic Mean

If x > 1, y>1...

Question

If $x > 1, y > 1, z > 1$ are in GP, then $\frac{1}{1 + l n x}, \frac{1}{1 + l n y}, \frac{1}{1 + l n z}$ are in

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Solution

The correct option is B HP
Let the common ratio of the GP be r. Then,
$y = x r$ and $z = x r^{2}$
$\Rightarrow l n y = l n x + l n r$ and $l n z = l n x + 2 l n r$
Let $A = 1 + l n x, D = l n r$
Then, $\frac{1}{1 + l n x} = \frac{1}{A}, \frac{1}{1 + l n y} = \frac{1}{1 + l n x + l n r} = \frac{1}{A + D}$
and $\frac{1}{1 + l n z} = \frac{1}{1 + l n x + 2 l n r} = \frac{1}{A + 2 D}$
Therefore, $\frac{1}{1 + l n x}, \frac{1}{1 + l n y}, \frac{1}{1 + l n z}$ are in HP

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If x>1,y>1,z>1 are in GP, then 11+ln x,11+ln y,11+ln z are in

The correct option is B HPLet the common ratio of the GP be r. Then, y=xr and z=xr2 ⇒ln y=ln x+ln r and ln z=ln x+2 ln r Let A=1+ln x, D=ln r Then, 11+ln x=1A,11+ln y=11+ln x+ln r=1A+D and 11+ln z=11+ln x+2 ln r=1A+2D Therefore, 11+ln x,11+ln y,11+ln z are in HP

If $x > 1, y > 1, z > 1$ are in GP, then $\frac{1}{1 + l n x}, \frac{1}{1 + l n y}, \frac{1}{1 + l n z}$ are in