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If \[{x^a} = {x^{b/2}}{z^{b/2}} = {z^c}\], then \[a,b,c\]are in
A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer
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Hint
By considering the reciprocals of the arithmetic progression that does not contain zero, a harmonic progression (HP) is defined as a sequence of real numbers. In mathematics, a set of numbers is referred to as an HP if the reciprocals of the terms are in AP. AP, GP, and HP stand for the average or mean of the series. Arithmetic Mean, Geometric Mean, and Harmonic Mean, respectively, are denoted by the letters AM, GM, and HM.
If the reciprocal of the terms is in AP, a series of numbers is referred to as a harmonic progression. If the reciprocal of the terms is in AP, a series of numbers is referred to as a harmonic progression.
Formula used:
\[\log (ab) = \log a + \log b\]
If \[a,b,c\] are in H.P. then
\[ \frac{2}{b} = \frac{1}{c} + \frac{1}{a}\]
Complete step-by-step solution
The given equation is
\[{x^a} = {x^{b/2}}{z^{b/2}}\]
Take log on both the sides, the equation becomes
\[a\log x = \frac{b}{2}\log x + \frac{b}{2}\log z\]
\[ = > \frac{b}{2}\log z = (a - \frac{b}{2})\log x\] ---(1)
Also, the given equation is \[{x^a} = {z^c}\]
On both sides, log value should be taken
\[a\log x = c\log z\]
\[ = > \log z = \frac{{a\log x}}{c}\] ---(2)
Substitute the value of log z in equation (1)
\[\frac{1}{{2c}} = \frac{1}{b} - \frac{1}{{2a}}\]
\[ = > \frac{1}{b} = \frac{1}{2}(\frac{1}{c} + \frac{1}{a})\]
\[ = > \frac{2}{b} = \frac{1}{c} + \frac{1}{a}\]
As \[a,b,c\] meet the requirements for H.P., they are thus in H.P.
Therefore, the correct option is C.
Note
The reciprocal of the nth term in the equivalent arithmetic progression is the nth term in a harmonic progression. In a geometric progression (GP), the common ratio is multiplied by the previous term to produce each succeeding term. If the reciprocal of the terms is in AP, a series of numbers is referred to as a harmonic progression. A progression is a pattern-following series of numbers. A series of real numbers known as a harmonic progression (HP) is created by taking the reciprocals of the arithmetic progression.