If x -1- y -1- z -1 are in G-P- then 1-1-In x- 1-1-In y - 1-1-In z are inA- A-P-B- H-P-C- G-P-D- None of these

The correct option is B H.P.x,y,x are in G.P. Hence y^2 xz ∴ 2 log y = log x + log z ⇒ 2(log y+1)=(1+log x)+(1+ log z) ⇒ 1 + log x, 1 + log y, 1 + log z are ...

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Byju's Answer

Standard XII

Mathematics

Harmonic Progression

If x >1, y >1...

Question

If $x > 1, y > 1, z > 1$ are in G.P., then $\frac{1}{1 + I n x}, \frac{1}{1 + I n y}, \frac{1}{1 + I n z}$ are in

A.P.

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H.P.

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G.P.

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None of these

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Solution

The correct option is B H.P.
x,y,x are in G.P. Hence $y^{2} - x z$
$∴ 2 log y = log x + log z$
$\Rightarrow 2 (log y + 1) = (1 + log x) + (1 + log z)$
$\Rightarrow 1 + log x, 1 + log y, 1 + log z$ are in A.P.
$\Rightarrow \frac{1}{1 + log x}, \frac{1}{1 + log y}, \frac{1}{1 + log z}$ are is H.P.

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If x>1,y>1,z>1 are in G.P., then 11+In x,11+In y,11+In z are in

The correct option is B H.P.x,y,x are in G.P. Hence y2−xz ∴2logy=logx+logz ⇒2(logy+1)=(1+logx)+(1+logz) ⇒1+logx,1+logy,1+logz are in A.P. ⇒11+logx,11+logy,11+logz are is H.P.

If $x > 1, y > 1, z > 1$ are in G.P., then $\frac{1}{1 + I n x}, \frac{1}{1 + I n y}, \frac{1}{1 + I n z}$ are in

The correct option is B H.P.
x,y,x are in G.P. Hence $y^{2} - x z$
$∴ 2 log y = log x + log z$
$\Rightarrow 2 (log y + 1) = (1 + log x) + (1 + log z)$
$\Rightarrow 1 + log x, 1 + log y, 1 + log z$ are in A.P.
$\Rightarrow \frac{1}{1 + log x}, \frac{1}{1 + log y}, \frac{1}{1 + log z}$ are is H.P.