We should accept the G.P. to be \[a,~ar,~a{{r}^{2}},~a{{r}^{3}},\text{ }\ldots \]

Then, at that point, as per the inquiry, we have

\[\begin{array}{*{35}{l}}

a~+~ar~+~a{{r}^{2}}~=\text{ }16\text{ }and~a{{r}^{3~}}+~a{{r}^{4}}~+~a{{r}^{5~}}=\text{ }128  \\

a~\left( 1\text{ }+~r~+~{{r}^{2}} \right)\text{ }=\text{ }16\text{ }\ldots \text{ }\left( 1 \right)\text{ }and,  \\

a{{r}^{3}}\left( 1\text{ }+~r~+~{{r}^{2}} \right)\text{ }=\text{ }128\text{ }\ldots \text{ }\left( 2 \right)  \\

\end{array}\]

Separating condition\[\left( 2 \right)\text{ }by\text{ }\left( 1 \right)\], we get

\[\begin{array}{*{35}{l}}

{{r}^{3}}~=\text{ }8  \\

r\text{ }=\text{ }2  \\

\end{array}\]

Presently, utilizing\[r\text{ }=\text{ }2\text{ }in\text{ }\left( 1 \right)\], we get

\[\begin{array}{*{35}{l}}

a\text{ }\left( 1\text{ }+\text{ }2\text{ }+\text{ }4 \right)\text{ }=\text{ }16  \\

a\text{ }\left( 7 \right)\text{ }=\text{ }16  \\

a\text{ }=\text{ }16/7  \\

\end{array}\]

Presently, the amount of terms is given as