From the question it is given that,

The geometric progression whose 4^th term \[{{a}_{4}}~=\text{ }54\]

The geometric progression whose 7^th term \[{{a}_{7}}~=\text{ }1458\]

We know that, a_n = ar^{n – 1}

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

{{a}_{4}}~=\text{ }a{{r}^{4\text{ }\text{ }1}}  \\

{{a}_{4}}~=\text{ }a{{r}^{3}}~=\text{ }54  \\

{{a}_{7}}~=\text{ }a{{r}^{6}}~=\text{ }1458  \\

\end{array}\]

By dividing both we get,

ar⁶/ar³ = 1458/54

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

{{r}^{6\text{ }\text{ }3}}~=\text{ }27  \\

{{r}^{3}}~=\text{ }{{3}^{3}}  \\

r\text{ }=\text{ }3  \\

\end{array}\]

To find out a, consider

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

a{{r}^{3}}~=\text{ }54  \\

a{{\left( 3 \right)}^{3}}~=\text{ }54  \\

a\text{ }=\text{ }54/27  \\

a\text{ }=\text{ }2  \\

\end{array}\]

Therefore, \[a\text{ }=\text{ }2,\text{ }r\text{ }=\text{ }3\]

So, G.P. is \[2,\text{ }6,\text{ }18,\text{ }54,\ldots \]