Given, the initial term of the G.P is an and the last term is
![\[b.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0efd9eafb3664c9e8b9cb1832ad3ce14_l3.png "Rendered by QuickLaTeX.com")
Hence,
The G.P.
![\[~a,~ar,~a{{r}^{2}},~a{{r}^{3}},\text{ }\ldots ~a{{r}^{n}}^{1},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-60d27e785dbe254a5c1a56be1b5bed70_l3.png "Rendered by QuickLaTeX.com")
is the place where r is the normal proportion.
Then, at that point,
![\[\begin{array}{*{35}{l}} b~=~a{{r}^{n}}^{1}~\text{ }\ldots \text{ }\left( 1 \right) \\ P~=\text{ }Product\text{ }of~n~terms \\ =\text{ }\left( a \right)\text{ }\left( ar \right)\text{ }\left( a{{r}^{2}} \right)\text{ }\ldots \text{ }\left( a{{r}^{n}}^{1} \right) \\ =\text{ }\left( a~\times ~a~\times \ldots a \right)\text{ }\left( r~\times ~{{r}^{2}}~\times \text{ }\ldots {{r}^{n}}^{1} \right) \\ =~{{a}^{n}}^{~}r{{~}^{1\text{ }+\text{ }2\text{ }+\ldots (}}{{^{n}}^{1)}}~\text{ }\ldots \text{ }\left( 2 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-398215f9dc55559c2d500a8935b170aa_l3.png "Rendered by QuickLaTeX.com")
Here,
![\[1,\text{ }2,\text{ }\ldots (n~\text{ }1)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6347ce984a46b4b07e57f263fb0a41f5_l3.png "Rendered by QuickLaTeX.com")
is an A.P.
Furthermore, the result of
![\[n\text{ }terms\text{ }P\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-178519ebf05e671cc5b97255861e47ee_l3.png "Rendered by QuickLaTeX.com")
is given by,
If the first and the nth term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.
If the first and the nth term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.