From the question it is given that,
The geometric progression whose 4th term \[{{a}_{4}}~=\text{ }54\]
The geometric progression whose 7th term \[{{a}_{7}}~=\text{ }1458\]
We know that, an = arn – 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 \[2,\text{ }6,\text{ }18,\text{ }54,\ldots \]