Let $7\stackrel{rtimes}{\overbrace{\mathrm{3.........3}}}7$ denotes $(r + 2)$ digit number where the first and last digits are 7 and the remaining r digits are 3. Consider the sum  $S=77+737+7337+\mathrm{....}+7\stackrel{98times}{\overbrace{\mathrm{3.........3}}}7$. If $S=\frac{7\stackrel{99times}{\overbrace{\mathrm{3.........3}}}7+a}{b}$ where a and b are natural numbers then sum of the digits of $a + b$ is

Let $S=77+737+7337+\mathrm{....}+\stackrel{98times}{\overbrace{\mathrm{73.........37}}}$  …….(1)
 $10S=770+7370+\mathrm{.........}+\mathrm{73........370}+\mathrm{73......370}$ ….. (2)
$$\begin{gathered} \left(1\right) – \left(2\right) \Rightarrow -9S=77-33-\mathrm{33......}-33-\stackrel{98times}{\overbrace{\mathrm{73.....370}}} \\ 9S=-77+33+33+\mathrm{........}+33+\mathrm{73......370} \\ =-77+33+33+\mathrm{..........}+33+\mathrm{73.....370}-33+33 \\ =-77+98\times 33+\stackrel{99times}{\overbrace{\mathrm{73......37}}} \\ S=\frac{3190+\stackrel{99times}{\overbrace{\mathrm{73.....37}}}}{9} \end{gathered}$$