Let
![\[a/r,\text{ }a,\text{ }ar\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-87f1add9e3ed72a23ab1fb4fac317af3_l3.png "Rendered by QuickLaTeX.com")
be the initial three terms of the
![\[G.P.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1dadd4e95aec87cc3adfa3fd45c98025_l3.png "Rendered by QuickLaTeX.com")
![\[\begin{array}{*{35}{l}} a/r\text{ }+\text{ }a\text{ }+\text{ }ar\text{ }=\text{ }39/10\text{ }\ldots \ldots \text{ }\left( 1 \right) \\ \left( a/r \right)\text{ }\left( a \right)\text{ }\left( ar \right)\text{ }=\text{ }1\text{ }\ldots \ldots ..\text{ }\left( 2 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b5b51f08ce9cbc5adbcdc928ccfbc43f_l3.png "Rendered by QuickLaTeX.com")
From
![\[\left( 2 \right),\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e5538a6dc85ee07fbc19ecab8b497a26_l3.png "Rendered by QuickLaTeX.com")
we have
![\[{{a}^{3}}~=\text{ }1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0bde425d4aa438637102cf79270c686a_l3.png "Rendered by QuickLaTeX.com")
Subsequently,
![\[a\text{ }=\text{ }1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-60007b777d643ba9eb870e9785edbbb6_l3.png "Rendered by QuickLaTeX.com")
[Considering genuine roots only]
Subbing the worth of
![\[a\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a3ab43dca3872e69a0c1be257c1ecec8_l3.png "Rendered by QuickLaTeX.com")
in
![\[\left( 1 \right),\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6ab88eefb94fe80e32a489967e1f88eb_l3.png "Rendered by QuickLaTeX.com")
we get
![\[\begin{array}{*{35}{l}} 1/r\text{ }+\text{ }1\text{ }+\text{ }r\text{ }=\text{ }39/10 \\ (1\text{ }+\text{ }r\text{ }+\text{ }{{r}^{2}})/r\text{ }=\text{ }39/10 \\ 10\text{ }+\text{ }10r\text{ }+\text{ }10{{r}^{2}}~=\text{ }39r \\ 10{{r}^{2}}~\text{ }29r\text{ }+\text{ }10\text{ }=\text{ }0 \\ 10{{r}^{2}}~\text{ }25r\text{ }\text{ }4r\text{ }+\text{ }10\text{ }=\text{ }0 \\ 5r\left( 2r\text{ }\text{ }5 \right)\text{ }\text{ }2\left( 2r\text{ }\text{ }5 \right)\text{ }=\text{ }0 \\ \left( 5r\text{ }\text{ }2 \right)\text{ }\left( 2r\text{ }\text{ }5 \right)\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-edfe9cc50b0004205fc8ce51d9c3eeaf_l3.png "Rendered by QuickLaTeX.com")
In this manner,
![\[r\text{ }=\text{ }2/5\text{ }or\text{ }5/2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-73f1e5661f91d613544ae760cbc2c664_l3.png "Rendered by QuickLaTeX.com")
In this manner, the three terms of the G.P. are
![\[5/2,\text{ }1\text{ }and\text{ }2/5.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-82748516d5626d3e28c289048850d18c_l3.png "Rendered by QuickLaTeX.com")