From the question it is given that,

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

{{T}_{9}}~=\text{ }1/7  \\

{{T}_{7}}~=\text{ }1/9  \\

\end{array}\]

Let us assume ‘a’ be the first term and ‘d’ be the common difference,

So, \[{{T}_{9}}~=\text{ }a\text{ }+\text{ }8d\text{ }=\text{ }1/7\] equation (i)

\[{{T}_{7}}~=\text{ }a\text{ }+\text{ }6d\text{ }=\text{ }1/9\] equation (ii)

Subtracting both equation (i) and equation (i),

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

\left( a\text{ }+\text{ }6d \right)\text{ }\text{ }\left( a\text{ }+\text{ }8d \right)\text{ }=\text{ }1/9\text{ }\text{ }1/7  \\

a\text{ }+\text{ }6d\text{ }\text{ }a\text{ }\text{ }8d\text{ }=\text{ }\left( 7\text{ }\text{ }9 \right)/63  \\

-2d\text{ }=\text{ }-2/63  \\

d\text{ }=\text{ }\left( -2/63 \right)\text{ }\times \text{ }\left( -1/2 \right)  \\

d\text{ }=\text{ }1/63  \\

\end{array}\]

now, substitute the value of d in equation (ii) to find out a, we get

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

a\text{ }=\text{ }1/9\text{ }\text{ }6/63  \\

a\text{ }=\text{ }\left( 7\text{ }\text{ }6 \right)/63  \\

a\text{ }=\text{ }1/63  \\

\end{array}\]

Therefore,

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

   {{T}_{63}}~=\text{ }a\text{ }+\text{ }62d  \\

   =\text{ }1/63\text{ }+\text{ }62\left( 1/63 \right)  \\

   =\text{ }1/63\text{ }+\text{ }62/63  \\

   =\text{ }\left( 1\text{ }+\text{ }62 \right)/63  \\

   =\text{ }63/63  \\

   =\text{ }1  \\

\end{array}\]