Question

If we have an expression as ${x^a} = {x^{\dfrac{b}{2}}}{z^{\dfrac{b}{2}}} = {z^c}$ , then $a,b,c$ are in
1. A.P
2. G.P
3. H.P
4. None of these

Answer 1

First, we need to know about the concept of Harmonic progression.
Harmonic progression is the reciprocal of arithmetic progression which can be represented as the sequence of $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},....$ where $a,b,c,...$ are need to be in the A.P
Harmonic progress is the reciprocal of the given arithmetic progression which is calculated in the form of $HP = \dfrac{1}{{[a + (n - 1)d]}}$ where $a$ is the first term and $d$ is a common difference and n is the number of AP.
A logarithm function or log operator is used when we have to deal with the powers and base of a number, to understand it better which is $\log {x^m} = m\log x$
Formula used:
$\log (ab) = \log a + \log b$, $\log (\dfrac{a}{b}) = \log a - \log b$

Complete step-by-step solution:
Since from the given that we have ${x^a} = {x^{\dfrac{b}{2}}}{z^{\dfrac{b}{2}}} = {z^c}$. This means ${x^a} = {x^{\dfrac{b}{2}}}{z^{\dfrac{b}{2}}}$ and also ${x^{\dfrac{b}{2}}}{z^{\dfrac{b}{2}}} = {z^c}$
Now taking logarithm on both the values we get ${x^a} = {(xz)^{\dfrac{b}{2}}} \Rightarrow a\log x = \dfrac{b}{2}(log(xz))$ and $x{z^{\dfrac{b}{2}}} = {z^c} \Rightarrow \dfrac{b}{2}(log(xz)) = clogz$
Since we know that $\log (ab) = \log a + \log b$ then we get $a\log x = \dfrac{b}{2}(logx + logz)$ and $\dfrac{b}{2}(logx + logz) = clogz$
Further solving with the subtraction, we have, $a\log x - \dfrac{b}{2}logx = \dfrac{b}{2}(logz)$ and $\dfrac{b}{2}(logx) = clogz - \dfrac{b}{2}\log z$
Now taking the common terms and divide them with the other side we get $\dfrac{{\log x}}{{\log z}} = \dfrac{{\dfrac{b}{2}}}{{(a - \dfrac{b}{2})}}$ and $\dfrac{{\log x}}{{\log z}} = \dfrac{{(c - \dfrac{b}{2})}}{{(\dfrac{b}{2})}}$
Since both the left side are the same a compare them, we have $\dfrac{{\dfrac{b}{2}}}{{(a - \dfrac{b}{2})}} = \dfrac{{(c - \dfrac{b}{2})}}{{(\dfrac{b}{2})}}$
Further solving with the cross multiplication, we have $\dfrac{{{b^2}}}{4} = (a - \dfrac{b}{2})(c - \dfrac{b}{2})$
$\dfrac{{{b^2}}}{4} = (a - \dfrac{b}{2})(c - \dfrac{b}{2}) \Rightarrow \dfrac{{{b^2}}}{4} = ac - \dfrac{{bc}}{2} - \dfrac{{ab}}{2} + \dfrac{{{b^2}}}{4}$
Canceling the common terms, we have $\dfrac{{{b^2}}}{4} = ac - \dfrac{{bc}}{2} - \dfrac{{ab}}{2} + \dfrac{{{b^2}}}{4} \Rightarrow 2ac = bc + ab$
Divide both side values with $abc$ then we get $2ac = bc + ab \Rightarrow \dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$
Hence, $a,b,c$ are in the harmonic progression, and thus the option 3) is correct.

Note: While talking about the A.P and G.P, we need to know about the concept of Arithmetic and Geometric progression.
An arithmetic progression can be represented by $a,(a + d),(a + 2d),(a + 3d),...$where $a$ is the first term and $d$ is a common difference.
A geometric progression can be given by $a,ar,a{r^2},....$ where $a$ is the first term and $r$ is a common ratio.
For GP with the common ratio the formula to be calculated $GP = \dfrac{a}{{r - 1}},r \ne 1,r < 0$and $GP = \dfrac{a}{{1 - r}},r \ne 1,r > 0$
Also, for A.P we have $AP = [a + (n - 1)d]$