How about we take
![\[a\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a3ab43dca3872e69a0c1be257c1ecec8_l3.png "Rendered by QuickLaTeX.com")
to be the initial term and
![\[r\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a1e6023f21df45b456a9cd5898649fe1_l3.png "Rendered by QuickLaTeX.com")
to be the normal proportion of the G.P.
Then, at that point, as indicated by the inquiry, we have
![\[\begin{array}{*{35}{l}} {{a}_{5}}~=~a~{{r}^{51~}}=~a~{{r}^{4}}~=~p~\ldots \text{ }\left( i \right) \\ {{a}_{8~}}=~a~{{r}^{81~}}=~a~{{r}^{7}}~=~q~\ldots \text{ }\left( ii \right) \\ {{a}_{11}}~=\text{ }a~{{r}^{111~}}=~a~{{r}^{10}}~=~s~\ldots \text{ }\left( iii \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-45ae04ade3c50afaf5b9d94903f30b33_l3.png "Rendered by QuickLaTeX.com")
Separating condition
![\[\left( ii \right)\text{ }by\text{ }\left( i \right),\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-34eacf4316170e913a1b7ab6ef584750_l3.png "Rendered by QuickLaTeX.com")
we get
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.