
If ${x^a} = {x^{\dfrac{b}{2}}}{z^{\dfrac{b}{2}}} = {z^c}$ then a, b, c are in
A. AP
B. HP
C. GP
D. None of these
Answer
233.1k+ views
Hint: Before we proceed with the problem, it is important to know the definitions of Progression, AP, GP, and HP. Before we proceed with the problem, it is important to know the definitions of Progression, AP, GP, and HP. We will also be using the logarithmic properties. Take the logarithmic function simultaneously on both equal functions. Solving and comparing the formed equation will give you the result.
Complete step by step Solution:
Given ${x^a} = {x^{\dfrac{b}{2}}}{z^{\dfrac{b}{2}}} = {z^c}$----(1)
$ \Rightarrow {x^a} = {(xz)^{\dfrac{b}{2}}}$
$ \Rightarrow {z^c} = {(xz)^{\dfrac{b}{2}}}$
Now taking the log on both sides of equation (1) we get,
$ \Rightarrow \log {x^a} = \log {(xz)^{\dfrac{b}{2}}} = \log {z^c}$
$ \Rightarrow a\log x = (\dfrac{b}{2})\log (xz) = c\log z$
$ \Rightarrow a\log x = (\dfrac{b}{2})\log x + (\dfrac{b}{2})\log z = c\log z$
$ \Rightarrow (a - \dfrac{b}{2})\log x = (\dfrac{b}{2})\log z$ and $(\dfrac{b}{2})\log x = (c - \dfrac{b}{2})\log z$
$ \Rightarrow \dfrac{{\log x}}{{\log z}} = \dfrac{{(\dfrac{b}{2})}}{{(a - \dfrac{b}{2})}}$ and $\dfrac{{\log x}}{{\log z}} = \dfrac{{(c - \dfrac{b}{2})}}{{(\dfrac{b}{2})}}$
Equating the LHS of the above equations we get,
$ \Rightarrow \dfrac{{(c - \dfrac{b}{2})}}{{(\dfrac{b}{2})}} = \dfrac{{(\dfrac{b}{2})}}{{(a - \dfrac{b}{2})}}$
$ \Rightarrow \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})$
$ \Rightarrow 2ac = bc + ab$
Divide by abc
$ \Rightarrow \dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$
So, a, b, and c are in HP.
Hence, the correct option is B.
Additional Information:
1. Progressions are numbers arranged in a particular order such that they form an expected order. By expected order, we mean that given some numbers, we can find the next numbers in the series.
2. Arithmetic Progression (AP) - A sequence is called an arithmetic progression if the difference between any two consecutive terms is always the same. In simple terms, it means that the next number in the series is calculated by adding a fixed number to the previous number in the series. This fixed number is a common difference.
3. Geometric Progression (GP) - A sequence is called a geometric progression if the ratio of any two consecutive terms is always the same. In simple terms, it means that the next number in the series is calculated by multiplying a fixed number by the previous number in the series. This fixed number is called the common ratio.
4. A sequence is called a harmonic progression if the reciprocal of the terms are in AP.
Note: While solving these types of questions we need to know about both progression and logarithmic properties. Care should be taken while applying property $\log ab = \log a + \log b$ and try to relate a, b, and c differently. Also, while dividing by abc take care none of them should be zero. If it is zero, we cannot divide.
Complete step by step Solution:
Given ${x^a} = {x^{\dfrac{b}{2}}}{z^{\dfrac{b}{2}}} = {z^c}$----(1)
$ \Rightarrow {x^a} = {(xz)^{\dfrac{b}{2}}}$
$ \Rightarrow {z^c} = {(xz)^{\dfrac{b}{2}}}$
Now taking the log on both sides of equation (1) we get,
$ \Rightarrow \log {x^a} = \log {(xz)^{\dfrac{b}{2}}} = \log {z^c}$
$ \Rightarrow a\log x = (\dfrac{b}{2})\log (xz) = c\log z$
$ \Rightarrow a\log x = (\dfrac{b}{2})\log x + (\dfrac{b}{2})\log z = c\log z$
$ \Rightarrow (a - \dfrac{b}{2})\log x = (\dfrac{b}{2})\log z$ and $(\dfrac{b}{2})\log x = (c - \dfrac{b}{2})\log z$
$ \Rightarrow \dfrac{{\log x}}{{\log z}} = \dfrac{{(\dfrac{b}{2})}}{{(a - \dfrac{b}{2})}}$ and $\dfrac{{\log x}}{{\log z}} = \dfrac{{(c - \dfrac{b}{2})}}{{(\dfrac{b}{2})}}$
Equating the LHS of the above equations we get,
$ \Rightarrow \dfrac{{(c - \dfrac{b}{2})}}{{(\dfrac{b}{2})}} = \dfrac{{(\dfrac{b}{2})}}{{(a - \dfrac{b}{2})}}$
$ \Rightarrow \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})$
$ \Rightarrow 2ac = bc + ab$
Divide by abc
$ \Rightarrow \dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$
So, a, b, and c are in HP.
Hence, the correct option is B.
Additional Information:
1. Progressions are numbers arranged in a particular order such that they form an expected order. By expected order, we mean that given some numbers, we can find the next numbers in the series.
2. Arithmetic Progression (AP) - A sequence is called an arithmetic progression if the difference between any two consecutive terms is always the same. In simple terms, it means that the next number in the series is calculated by adding a fixed number to the previous number in the series. This fixed number is a common difference.
3. Geometric Progression (GP) - A sequence is called a geometric progression if the ratio of any two consecutive terms is always the same. In simple terms, it means that the next number in the series is calculated by multiplying a fixed number by the previous number in the series. This fixed number is called the common ratio.
4. A sequence is called a harmonic progression if the reciprocal of the terms are in AP.
Note: While solving these types of questions we need to know about both progression and logarithmic properties. Care should be taken while applying property $\log ab = \log a + \log b$ and try to relate a, b, and c differently. Also, while dividing by abc take care none of them should be zero. If it is zero, we cannot divide.
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