Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

How about we accept

    \[{{G}_{1}}~and~{{G}_{2}}\]

to be two numbers somewhere in the range of

    \[3\text{ }and\text{ }81\]

with the end goal that the series

    \[3,~{{G}_{1}},~{{G}_{2}},\text{ }81~\]

frames a G.P.

Furthermore, let

    \[a\]

be the initial term and

    \[r\]

be the normal proportion of the G.P.

Presently, we have the

    \[~{{1}^{st}}~term\text{ }as\text{ }3\text{ }and\text{ }the\text{ }{{4}^{th}}~term\text{ }as\text{ }81.\]

    \[\begin{array}{*{35}{l}} 81\text{ }=\text{ }\left( 3 \right)~{{(r)}^{3}}  \\ {{r}^{3}}~=\text{ }27  \\ \therefore ~r~=\text{ }3\text{ }\left( Taking\text{ }real\text{ }roots\text{ }only \right)  \\ For~r~=\text{ }3,  \\ {{G}_{1}}~=~ar~=\text{ }\left( 3 \right)\text{ }\left( 3 \right)\text{ }=\text{ }9  \\ {{G}_{2}}~=~a{{r}^{2}}~=\text{ }\left( 3 \right)\text{ }{{\left( 3 \right)}^{2}}~=\text{ }27  \\ \end{array}\]

Hence, the two numbers which can be embedded somewhere in the range of

    \[3\text{ }and\text{ }81\]

so the subsequent arrangement turns into a G.P are

    \[9\text{ }and\text{ }27.\]