If xa-xb - 2 zb - 2-zc- then a- b- c are inA- A-P-B- G-PC- H-P-D- None

The correct option is C H.P. nbsp;Let x^a=(xz)^b/2=z^c=t ⇒ x=t^1/a, xz=t^2/b, z=t^1/c ∵ xz=t^2/b ⇒ t^1/a t^1/c=t^2/b ⇒1a+1c=2b So, a,b,c are in H.P. ...

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Standard XII

Mathematics

Harmonic Mean

If xa=xb / 2 ...

Question

If $x^{a} = x^{b / 2} z^{b / 2} = z^{c}$ , then $a, b, c$ are in

A . P .

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G . P .

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H . P .

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None

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Solution

The correct option is C $H . P .$
Let $x^{a} = (x z)^{b / 2} = z^{c} = t$
$\Rightarrow x = t^{1 / a}, x z = t^{2 / b}, z = t^{1 / c} ∵ x z = t^{2 / b} \Rightarrow t^{1 / a} t^{1 / c} = t^{2 / b} \Rightarrow \frac{1}{a} + \frac{1}{c} = \frac{2}{b}$

So, $a, b, c$ are in $H . P .$

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If xa=xb/2zb/2=zc, then a,b,c are in

The correct option is C H.P. Let xa=(xz)b/2=zc=t ⇒x=t1/a, xz=t2/b, z=t1/c∵xz=t2/b⇒t1/a t1/c=t2/b⇒1a+1c=2b So, a,b,c are in H.P.

If $x^{a} = x^{b / 2} z^{b / 2} = z^{c}$ , then $a, b, c$ are in

The correct option is C $H . P .$
Let $x^{a} = (x z)^{b / 2} = z^{c} = t$
$\Rightarrow x = t^{1 / a}, x z = t^{2 / b}, z = t^{1 / c} ∵ x z = t^{2 / b} \Rightarrow t^{1 / a} t^{1 / c} = t^{2 / b} \Rightarrow \frac{1}{a} + \frac{1}{c} = \frac{2}{b}$

So, $a, b, c$ are in $H . P .$