Solveeit Logo

Question

Question: If \(a , b, c\) are in G.P. then \({\log _a}x,\,{\log _b}x,\,{\log _c}x\) are in A) A.P. B) G.P....

If a,b,ca , b, c are in G.P. then logax,logbx,logcx{\log _a}x,\,{\log _b}x,\,{\log _c}x are in
A) A.P.
B) G.P.
C) H.P.
D) None of the above

Explanation

Solution

In order to solve this question, you have to know basic details about Geometric Progression(G.P). In this question a,b,ca, b, c are in G.P. which means the square of second term(bb) is the product of first term(aa) and third term(cc) i.e., b2=ac{b^2} = ac. We use this logic og G.P. along with logarithmic formulas to get the required result.

Formula used:
- logba=alogb\log b^a = a \log b
- logab=loga+logb\log ab = log a + log b
- logyx=logxlogy{\log_y} x = \dfrac{\log x}{\log y}
- If three terms a,b,ca,b,c are in A.P. then it should satisfy 2b=a+c2b=a+c
- If three terms a,b,ca,b,c are in G.P. then b2=abb^2=ab
- A sequence of numbers is called a harmonic progression if the reciprocal of the terms are in AP. If three terms a,b,ca,b,c are in H.P. 1a,1b,1c\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c} are in A.P, then 2b=1a+1b\dfrac{2}{b}= \dfrac{1}{a}+\dfrac{1}{b}.

Complete step by step answer:
We are given that the variables a,b,ca, b, c are in geometric progression.
We know that when variables are in geometric progress then this means the square of the second term is the product of the first term and third term.
Here the first term is aa, second term is bb and third one is cc.
Which means b2=ac{b^2} = ac .
And we have to find logax,logbx,logcx{\log _a}x,\,{\log _b}x,\,{\log _c}x are in which we progress.
b2=ac\Rightarrow {b^2} = ac
Now, apply logarithm to the above equation, we will get,
By using logarithm properties logba=alogb{\log b^a} = a\log b and logab=loga+logb\log ab= \log a+ \log b
We get,
2logb=loga+logc\Rightarrow 2\log b = \log a + \log c
Divide by logx\log x in denominator we will get,
2logblogx=logalogx+logclogx\Rightarrow 2\dfrac{{\log b}}{{\log x}} = \dfrac{{\log a}}{{\log x}} + \dfrac{{\log c}}{{\log x}}
Now, apply basic steps of logarithm logmlogn=lognm\dfrac{{\log m}}{{\log n}} = {\log _n}m
And we will find something like the following equation,
2logxb=logxa+logxc\Rightarrow 2{\log _x}b = {\log _x}a + {\log _x}c
Take logarithm in denominator.
By using lognm=1logmn{\log _n}m = \dfrac{1}{{{{\log }_m}n}} . We will get,
2logax=1logbx+1logcx\Rightarrow \dfrac{2}{{{{\log }_a}x}} = \dfrac{1}{{{{\log }_b}x}} + \dfrac{1}{{{{\log }_c}x}}
As for sequence being in HP the term will be relation to each other like,
2b=1a+1b\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{b}
And this is the equation of H.P.
Therefore, logax,logbx,logcx{\log _a}x,\,{\log _b}x,\,{\log _c}x are in H.P. So, option (C) is correct.

Note:
- Now, let’s see a few things about Harmonic Progression (H.P.). It is defined as a sequence of real numbers which is determined by taking the reciprocals of the arithmetic progression that does not contain 0.
- In harmonic progression, any term in the sequence is considered as the harmonic means of its two neighbors.
- Harmonic mean is calculated as the reciprocal of the arithmetic mean of the reciprocals.
Harmonic Progression (H.P.) formula = 1a+(n1)d\dfrac{1}{{a + (n - 1)d}} .