Question

If \[a\], \[b\], and \[c\] are in G.P., then prove that \[\log a\], \[\log b\], and \[\log c\] are in A.P.

Answer 1

Hint: Here, we need to prove that \[\log a\], \[\log b\], and \[\log c\] are in A.P. We will use the formula for \[{n^{{\rm{th}}}}\] term of a G.P. to form an equation. Then, we will take the logarithm on both the sides. Finally, we will use the rules of logarithms and simplify the equation to prove that \[\log a\], \[\log b\], and \[\log c\] are in A.P. Any three numbers \[x\], \[y\], and \[z\] are in A.P. if \[x + z = 2y\].

Formula Used:
We will use the following formulas:
1.The \[{n^{{\rm{th}}}}\] term of a G.P. is given by the formula \[{a_n} = a{r^{n - 1}}\], where \[a\] is the first term and \[r\] is the common ratio.
2.\[\log {x^n} = n\log x\].
3.\[\log \left( {xy} \right) = \log x + \log y\].

Complete step-by-step answer:
First, we will use the formula for the \[{n^{{\rm{th}}}}\] term of a G.P.
It is given that \[a\], \[b\], and \[c\] are in G.P.
Therefore, the first term of the G.P. is \[a\], \[b\] is the second term of the G.P., and \[c\] is the third term of the G.P.
Substituting \[n = 2\] in the formula for \[{n^{{\rm{th}}}}\] term of a G.P., \[{a_n} = a{r^{n - 1}}\], we get
\[ \Rightarrow {a_2} = a{r^{2 - 1}}\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow {a_2} = a{r^1}\\ \Rightarrow {a_2} = ar\\ \Rightarrow b = ar\end{array}\]
Substituting \[n = 3\] in the formula for \[{n^{{\rm{th}}}}\] term of a G.P., \[{a_n} = a{r^{n - 1}}\], we get
\[ \Rightarrow {a_3} = a{r^{3 - 1}}\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow {a_3} = a{r^2}\\ \Rightarrow c = a{r^2}\end{array}\]
Multiplying both sides by the first term \[a\], we get
\[\begin{array}{l} \Rightarrow c \times a = a{r^2} \times a\\ \Rightarrow ac = {a^2}{r^2}\end{array}\]
Rewriting the expression, we get
\[ \Rightarrow ac = {\left( {ar} \right)^2}\]
Substituting \[b = ar\] in the equation, we get
\[ \Rightarrow ac = {b^2}\]
Now, we will prove that \[\log a\], \[\log b\], and \[\log c\] are in A.P.
We know that if \[x = y\], then \[\log x = \log y\].
Therefore, since \[ac = {b^2}\], we get
\[ \Rightarrow \log \left( {ac} \right) = \log \left( {{b^2}} \right)\]
The logarithm of a number raised to a power can be written as the product of the power, and the logarithm of the number. This can be written as \[\log {x^n} = n\log x\].
Substituting \[x = b\] and \[n = 2\] in the rule of logarithms \[\log {x^n} = n\log x\], we get
\[ \Rightarrow \log {b^2} = 2\log b\]
Substituting \[\log {b^2} = 2\log b\] in the equation \[\log \left( {ac} \right) = \log \left( {{b^2}} \right)\], we get
\[ \Rightarrow \log \left( {ac} \right) = 2\log b\]
The logarithm of the product of two numbers can be written as the sum of the logarithms of the two numbers. This can be written as \[\log \left( {xy} \right) = \log x + \log y\]. Here, the base of the terms need to be equal.
Substituting \[x = a\] and \[y = c\] in the rule of logarithms \[\log \left( {xy} \right) = \log x + \log y\], we get
\[ \Rightarrow \log \left( {ac} \right) = \log a + \log c\]
Substituting \[\log \left( {ac} \right) = \log a + \log c\] in the equation \[\log \left( {ac} \right) = 2\log b\], we get
\[ \Rightarrow \log a + \log c = 2\log b\]
We know that three numbers \[x\], \[y\], and \[z\] are in A.P. if \[x + z = 2y\].
Therefore, since \[\log a + \log c = 2\log b\], the numbers \[\log a\], \[\log b\], and \[\log c\] are in A.P.
Hence, we have proved that \[\log a\], \[\log b\], and \[\log c\] are in A.P.

Note: An arithmetic progression is a series of numbers in which each successive number is the sum of the previous number and a fixed difference. The fixed difference is called the common difference.
A geometric progression is a series of numbers in which each successive number is the product of the previous number and a fixed ratio. The fixed ratio is called the common ratio.
Here, we need to keep in mind different rules of logarithms to simplify the equation. If we don’t know the rules we might make a mistake by writing \[\log \left( {xy} \right) = \log x - \log y\] instead of \[\log \left( {xy} \right) = \log x + \log y\], which will incur the wrong answer.