The correct option is B GPGiven series 1c,(1ca)^12,1a Lets consider a G.P of elements A,B,C #160; ∴ Geo.mean ⇒ B^2=AC Comparing it with g ...

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1/c,1/ca12,1/...

Question

$\frac{1}{c}, (\frac{1}{c a})^{\frac{1}{2}}, \frac{1}{a}$ is in

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Solution

The correct option is B GP
Given series

$\frac{1}{c}, {(\frac{1}{c a})}^{\frac{1}{2}}, \frac{1}{a}$

Lets consider a G.P of elements $A, B, C$

$∴$ Geo.mean $\Rightarrow B^{2} = A C$

Comparing it with given series.

$A = \frac{1}{c} B = {(\frac{1}{c a})}^{\frac{1}{2}}, C = \frac{1}{a}$

$∴ B^{2} = {(\frac{1}{c a})}^{\frac{1}{2} \times 2}$

$= \frac{1}{c a}$ ........(1)

$A C = \frac{1}{c} \times \frac{1}{a} = \frac{1}{c a}$ ..............(ii)

$∴ (i) = (i i)$

$∴ B^{2} = A C$ So given series is in G.P

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1c,(1ca)12,1a is in

The correct option is B GPGiven series1c,(1ca)12,1aLets consider a G.P of elements A,B,C ∴ Geo.mean ⇒B2=ACComparing it with given series.A=1cB=(1ca)12,C=1a∴B2=(1ca)12×2 =1ca........(1)AC=1c×1a=1ca..............(ii)∴(i)=(ii)∴B2=AC So given series is in G.P

$\frac{1}{c}, (\frac{1}{c a})^{\frac{1}{2}}, \frac{1}{a}$ is in