The Geometric mean \(G\) of any two numbers \(a\) and \(b\) is given by \(\sqrt {ab} \) where the terms \(a\), \(G\), \(b\) are in \(G.P.\) If \({a_1}\), \({a_2}\), \({a_3}\), ………, \({a_n}\) are \(n\) numbers, then the Geometric mean \(G\), of these numbers is given by  \(G = {\left( {{a_1}{a_2}{a_3}....{a_n}} \right)^{1/n}}\)

The \(n\) number \({G_1}\), \({G_2}\), …… \({G_n}\) are said to be geometric means between \(a\) and \(b\) if \(a\), \({G_1}\), \({G_2}\), ….., \({G_n}\), \(b\) is a \(G.P.\) Here \(a\) is the first term and \(b\) is the \({\left( {n + {\rm{ }}2} \right)^{th}}\) term of the \(G.P.\) if \(r\) is the common ratio of this \(G.P.\), then 

\(b = a{r^{n + 1}}\) gives \(r = {\left( {\frac{b}{a}} \right)^{1/\,(n\, + \,1)}}\)

Thus, \({G_1} = ar,{G_{2\;}} = a{r^2},{\rm{ }} \ldots  \ldots .,{G_n} = a{r^n}\)

Note that \({G_1}{G_2}, \ldots  \ldots ..{G_n} = {\rm{ }}{\left( {ab} \right)^n}^{/2}\)

Example 13: The product of three numbers in G.P. is 27 and the sum of their product taken in pairs is 39, find the numbers.

Solution: Let the three numbers be \(\frac{a}{r},\,\,a\,\,\) and ar. According to the given conditions, we have

\(\frac{a}{r}.a.(ar) = 27\) \( \Rightarrow {a^3} = {\rm{ }}27\) that means a = 3

and \(\left( {\frac{a}{r}.a} \right) + (a.ar) + \left( {\frac{a}{r}.ar} \right) = 39\)

i.e. \({a^2}\left( {\frac{1}{r} + r + 1} \right) = 39\) i.e. \(9\left( {{r^{2\;}} + {\rm{ }}r{\rm{ }} + {\rm{ }}1} \right){\rm{ }} = {\rm{ }}39r\)

i.e. \(9{r^{2\;}}-{\rm{ }}30r + {\rm{ }}9{\rm{ }} = {\rm{ }}0 \Rightarrow \) \(r = 3,\,\,\,\frac{1}{3}\)

Hence, the numbers are \(1, 3, 9\) or \(9, 3,1\) 

Example 14: If \({\log_x}a,{\rm{ }}{a^{x/2}}\) and \({\log_b}x\) are in G.P., then find the value of \(x\).

Solution: According to the given condition, we have  \({\left( {{a^x}^{/2}} \right)^2} = \left( {{\log_x}a} \right)\left( {{\log_b}x} \right)\)

i.e. \({a^x} = \frac{{{{\log}_x}a}}{{{{\log }_x}b}}\),  which gives, \(x = {\log_a}\left( {{\log_b}a} \right)\)

Example 15: Find the sum up to 30 terms of the series \(7 + 77 + 777 + …….. 30\) terms.

Solution: Suppose the sum is \(S\), then

\(S = 7 + 77 + 777 + …….. 30\) terms

\(S = \frac{7}{9}\) [ \(9 +99 + 999 + …….30\) terms].

\(S = \frac{7}{9}\left[ {(10 - 1) + ({{10}^2} - 1) + ({{10}^3} - 1) + ....30\;terms} \right]\)

\(S = \frac{7}{9}\left[ {\frac{{10\left( {{{10}^{30}} - 1} \right)}}{{(10 - 1)}} - 30} \right]\) = \(\frac{7}{9}\left[ {\frac{{10\left( {{{10}^{30}} - 1} \right)}}{9} - 30} \right]\)