If the sum of the series 1 1 ·(1+ d)+ 1 (1+ d)(1+2 ~d)+…+ 1 (1+9 ~d)(1+10 ~d) is equal to 5 , then 50 ~d is equal to :

$\begin{aligned} & \frac{1}{1 \cdot(1+\mathrm{d})}+\frac{1}{(1+\mathrm{d})(1+2 \mathrm{~d})}+\ldots \ldots \ldots \ldots . . \\ & \frac{1}{(1+9 \mathrm{~d})(1+10 \mathrm{~d})}=5\end{aligned}$​1⋅(1+d)1​+(1+d)(1+2 d)1​+…………..(1+9 d)(1+10 d)1​=5​ $\begin{aligned} & \frac{1}{d}\left[\frac{(1+d)-1}{1 \cdot(1+d)}+\frac{(1+2 d)-(1-d)}{(1+d)(1+2 d)}\right]+\ldots \ldots \ldots \ldots . . . \\ & \frac{(1+10 d)-(1+9 d)}{(1+9 d)(1+10 d)}=5 \\ & \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}\right)+\ldots \ldots \ldots \ldots . .\right. \\ & \left.\left(\frac{1}{1+9 d}-\frac{1}{1+10 d}\right)\right]=5 \\ & \frac{1}{d}\left[1-\frac{1}{(1+10 d)}\right]=5 \\ & \frac{10 d}{1+10 d}=5 d \\ & 50 d=5\end{aligned}$​d1​[1⋅(1+d)(1+d)−1​+(1+d)(1+2d)(1+2d)−(1−d)​]+…………...(1+9d)(1+10d)(1+10d)−(1+9d)​=5d1​[(1−1+d1​)+(1+d1​−1+2d1​)+…………..(1+9d1​−1+10d1​)]=5d1​[1−(1+10d)1​]=51+10d10d​=5d50d=5​