Solution

\[ \begin{aligned} &\color{magenta}\textbf{First Year} \\[6pt] P &= P \\ R &= 10\% \\ T &= 1\ \text{year} \\[8pt] \text{1st year } (SI) &= \frac{P \times 10 \times 1}{100} \\[6pt] &= \frac{10P}{100} \\[6pt] &= \color{green} \frac{P}{10} \\[6pt] \text{1st year } (A) &= P + SI \\[6pt] &= P + \frac{P}{10} \\[6pt] &= \frac{11P}{10} \end{aligned} \]\[ \begin{aligned} &\color{magenta}\textbf{Second Year} \\[6pt] P &= \frac{11P}{10} \\[6pt] R &= 10\% \\ T &= 1\ \text{year} \\[8pt] \text{2nd year } (SI) &= \frac{11P \times 10 \times 1}{10 \times 100} \\[6pt] &= \frac{11P}{100} \\[6pt] \text{2nd year } (A) &= \frac{11P}{10} + \frac{11P}{100} \\[6pt] &= \frac{110P + 11P}{100} \\[6pt] &= \frac{121P}{100} \end{aligned} \]\[ \begin{aligned} &\color{magenta}\textbf{Third Year} \\[6pt] P &= \frac{121P}{100} \\[6pt] R &= 10\% \\ T &= 1\ \text{year} \\[8pt] \text{3rd year } (SI) &= \frac{121P \times 10 \times 1}{100 \times 100} \\[8pt] &=\color{green} \frac{121P}{1000} \\ \\ \end{aligned} \]\[ \begin{aligned} \color{magenta}\textbf{Difference}& = 1105 \\[6pt] \text{ 3rd year (SI) - 1st year (SI) } &= 1105 \\[6pt] \frac{121P}{1000} - \frac{P}{10} & = 1105 \\[6pt] \frac{121P - 100P}{1000} &= 1105 \\[6pt] \frac{21P}{1000} &= 1105 \\[6pt] P &= \frac{1105 \times 1000}{21} \\[6pt] P &= \frac{1105 \times 1000}{21} \\[6pt] P &= \text{₹ }52{,}619.04 \end{aligned} \]

Answer Sum invested \( = \color{red}\text{₹ }52{,}619.04 \)