If the sum of the series $$\frac{1}{1 \cdot (1+d)} + \frac{1}{(1+d)(1+2d)} + \ldots + \frac{1}{(1+9d)(1+10d)}$$ is equal to $$5$$, then $$50d$$ is equal to :

The series is $$\sum_{k=0}^{9} \frac{1}{(1+kd)(1+(k+1)d)}$$.

Using partial fractions: $$\frac{1}{(1+kd)(1+(k+1)d)} = \frac{1}{d}\left(\frac{1}{1+kd} - \frac{1}{1+(k+1)d}\right)$$.

This is a telescoping series:

$$\frac{1}{d}\left(\frac{1}{1} - \frac{1}{1+10d}\right) = \frac{1}{d} \cdot \frac{10d}{1+10d} = \frac{10}{1+10d}$$

Setting equal to 5: $$\frac{10}{1+10d} = 5 \implies 1+10d = 2 \implies d = \frac{1}{10}$$.

$$50d = 50 \times \frac{1}{10} = 5$$.

The correct answer is Option 2: 5.