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

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{1}{1 \cdot(1+\mathrm{d})}+\frac{1}{(1+\mathrm{d})(1+2 \mathrm{~d})}+\ldots \ldots \ldots \ldots . . \ & \frac{1}{(1+9 \mathrm{~d})(1+10 \mathrm{~d})}=5\end{aligned}$ $\begin{aligned} & \frac{1}{d}\left[\frac{(1+d)-1}{1 \cdot(1+d)}+\frac{(1+2 d)-(1-d)}{(1+d)(1+2 d)}ight]+\ldots \ldots \ldots \ldots . . . \ & \frac{(1+10 d)-(1+9 d)}{(1+9 d)(1+10 d)}=5 \ & \frac{1}{d}\left[\left(1-\frac{1}{1+d}ight)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}ight)+\ldots \ldots \ldots \ldots . .ight. \ & \left.\left(\frac{1}{1+9 d}-\frac{1}{1+10 d}ight)ight]=5 \ & \frac{1}{d}\left[1-\frac{1}{(1+10 d)}ight]=5 \ & \frac{10 d}{1+10 d}=5 d \ & 50 d=5\end{aligned}$

If the sum of the series $\frac{1}{1 \cdot(1+\mathrm{d})}+\frac{1}{(1+\mathrm{d})(1+2…

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