If sum of the infinite G.P. $p+1+\frac{1}{p}+\frac{1}{p^2}+\dots ,(p>2)$ is 49/6, then sum of the 3rd term and the 4th term of the G.P. is

For $p > 1,$ $\frac{49}{6}=p+1+\frac{1}{p}+\frac{1}{p^2}+\dots =\frac{p}{1-1/p}$

            $\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{49}{6}=\frac{p^2}{p-1}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6p^2-49p+49=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6p^2-42p-7p+49=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(6p-7)(p-7)=0\end{array}$

As                $p > 2, p = 7$

Thus,         $\begin{array}{l}{a}_3+a_4=p{\left(\frac{1}{p}\right)}^2+p{\left(\frac{1}{p}\right)}^3\\ =\frac{1}{7}+\frac{1}{49}=\frac{8}{49}\end{array}$