If x 1, y 1, z 1 are in G.P. then (1/1+log x), (1/1+log y), (1/1+log z) are in :

Question

If x > 1, y > 1, z > 1 are in G.P. then (1/1+log x), (1/1+log y), (1/1+log z) are in : (A) A.P (B) H.P (C) G.P (D) None of these

Tardigrade Admin · Accepted Answer

x, y, z are in GP ⇒ y 2= xz Taking log on both sides beginarrayl ⇒ 2 ln y= ln x+ ln z ⇒ 2(1+ ln y)=(1+ ln x)+(1+ ln z) endarray i.e 1+ ln x , 1+ ln y , 1+ ln z are in AP ∴ (1/1+ ln x), (1/1+ ln y), (1/1+ ln z) are in HP.

Q. If $x > 1, y > 1, z > 1$ are in $G . P .$ then $\frac{1}{1 + l o g x}, \frac{1}{1 + l o g y}, \frac{1}{1 + l o g z}$ are in :

$A . P$

$H . P$

$G . P$

None of these

$x, y, z$ are in GP
$\Rightarrow y^{2} = x z$
Taking log on both sides
$\Rightarrow 2 ln y = ln x + ln z \Rightarrow 2 (1 + ln y) = (1 + ln x) + (1 + ln z)$
i.e $1 + ln x, 1 + ln y, 1 + ln z$ are in $A P$
$∴ \frac{1}{1 + l n x}, \frac{1}{1 + l n y}, \frac{1}{1 + l n z}$ are in HP.

Q. If x>1,y>1,z>1 are in G.P. then 1+logx1​,1+logy1​,1+logz1​ are in :

A.P

H.P

G.P