Let $f\left(x\right)=x^2+x$ for all real x, there exists positive integers m and n, and distinct non-zero real numbers p and q such the $f\left(p\right)=f\left(q\right)=m+\sqrt{n}$  and if $f\left(\frac{1}{p}\right)+f\left(\frac{1}{q}\right)=\frac{1}{16}$  then the value of sum of digits in $100m+n$  is

Let $a=m+\sqrt{n}\Rightarrow \begin{array}{l}f\left(p\right)=a\\ f\left(q\right)=a\end{array}\}\Rightarrow x^2+x-a=0$ has root p ,q
$$\begin{gathered} \therefore f\left(\frac{1}{p}\right)+f\left(\frac{1}{q}\right)=\frac{1}{16}\Rightarrow \frac{1}{p^2}+\frac{1}{q^2}+\frac{1}{p}+\frac{1}{q}=\frac{1}{16}\Rightarrow \frac{p^2+q^2}{{\left(pq\right)}^2}+\frac{p+q}{pq}=\frac{1}{16} \\ \Rightarrow \frac{1+2a}{a^2}+\frac{1}{a}=\frac{1}{16}\Rightarrow \frac{1}{a^2}+\frac{3}{a}-\frac{1}{16}=0 \\ \begin{array}{l}=16+48a-a^2=0\Rightarrow a^2-48a-16=0\\ \Rightarrow a=\frac{48\pm \sqrt{2304+64}}{2}\Rightarrow a=24\pm \sqrt{592}\\ \Rightarrow a=24+\sqrt{592}\\ \therefore \begin{array}{l}m=24\\ n=592\end{array}\}\Rightarrow 100m+n=2992\end{array} \end{gathered}$$