Given series: \[\surd 2,\text{ }2,\text{ }2\surd 2,\text{ }\ldots \ldots \text{ },\text{ }32\]

Here,

\[a\text{ }=\text{ }\surd 2\]

\[r\text{ }=\text{ }2/\text{ }\surd 2\text{ }=\text{ }\surd 2\]

And, the \[last\text{ }term\text{ }\left( l \right)\text{ }=\text{ }32\]

\[l\text{ }=\text{ }{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}~=\text{ }32\]

\[\left( \surd 2 \right){{\left( \text{ }\surd 2 \right)}^{n\text{ }-\text{ }1}}~=\text{ }32\]

\[{{\left( \surd 2 \right)}^{n~}}=\text{ }32\]

\[{{\left( \surd 2 \right)}^{n~}}=\text{ }{{\left( 2 \right)}^{5}}~=\text{ }{{\left( \surd 2 \right)}^{10}}\]

On equating, we get

\[n\text{ }=\text{ }10\]

So, \[{{7}^{th}}\] term from the end is \[{{\left( 10\text{ }-\text{ }7\text{ }+\text{ }1 \right)}^{th~}}term.\]

i.e. \[{{4}^{th}}~\]term of the G.P.

Hence,

\[{{t}_{4}}~=\text{ }\left( \surd 2 \right){{\left( \surd 2 \right)}^{4\text{ }-\text{ }1}}~=\text{ }\left( \surd 2 \right){{\left( \surd 2 \right)}^{3}}~\]

\[=\text{ }\left( \surd 2 \right)\text{ }x\text{ }2\surd 2\text{ }=\text{ }4\]