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

The correct option is C H.P.x, y, z are in G.P. ⇒log x, log y, log z are in A.P. ⇒ (1+log x), (1+log y), (1+log z) are in A.P. Therefore, 11+log x, 11+log y, 1 ...

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Byju's Answer

Standard XI

Mathematics

Logarithmic Inequalities

If X , y , z ...

Question

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

A.P.

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G.P.

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H.P.

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None of these

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Solution

The correct option is C H.P.
$x, y, z$ are in G.P.
$\Rightarrow log x, log y, log z$ are in A.P.
$\Rightarrow (1 + log x), (1 + log y), (1 + log z)$ are in A.P.
Therefore,
$\frac{1}{1 + log x}, \frac{1}{1 + log y}, \frac{1}{1 + log z}$ are in H.P.

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If x,y,z are in G.P., where x,y,z>1, then 11+logx,11+logy,11+logz are in

The correct option is C H.P.x,y,z are in G.P. ⇒logx,logy,logz are in A.P. ⇒(1+logx),(1+logy),(1+logz) are in A.P. Therefore, 11+logx,11+logy,11+logz are in H.P.

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

The correct option is C H.P.
$x, y, z$ are in G.P.
$\Rightarrow log x, log y, log z$ are in A.P.
$\Rightarrow (1 + log x), (1 + log y), (1 + log z)$ are in A.P.
Therefore,
$\frac{1}{1 + log x}, \frac{1}{1 + log y}, \frac{1}{1 + log z}$ are in H.P.