Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

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Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

Consider

$a$

to be the initial term and

$r$

to be the normal proportion of the G.P.

Then, at that point,

${{a}_{1}}~=~a,~{{a}_{2}}~=~ar,~{{a}_{3}}~=~a{{r}^{2}},~{{a}_{4}}~=~a{{r}^{3}}$

From the inquiry, we have

$\begin{array}{*{35}{l}} {{a}_{3}}~=~{{a}_{1}}~+\text{ }9 \\ a{{r}^{2}}~=~a~+\text{ }9\text{ }\ldots \text{ }\left( i \right) \\ {{a}_{2}}~=~{{a}_{4}}~+\text{ }18 \\ ar~=~a{{r}^{3}}~+\text{ }18\text{ }\ldots \text{ }\left( ii \right) \\ \end{array}$

Thus, from

$\left( 1 \right)\text{ }and\text{ }\left( 2 \right),$

we get

$\begin{array}{*{35}{l}} a\left( {{r}^{2}}~\text{ }1 \right)\text{ }=\text{ }9\text{ }\ldots \text{ }\left( iii \right) \\ ar~\left( 1~{{r}^{2}} \right)\text{ }=\text{ }18\text{ }\ldots \text{ }\left( iv \right) \\ \end{array}$