JEE MAIN - Mathematics (2024 - 9th April Morning Shift - No. 7)

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 :
5
10
15
20

Explanation

$$\frac{1}{1 \cdot(1+d)}+\frac{1}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}=5$$

Multiply and divide by $$d$$

$$\begin{aligned} & \frac{1}{d}\left[\frac{d}{1 \times(1+d)}+\frac{d}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}\right]=5 \\ & \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}\right)+\ldots+\left(\frac{1}{1+9 d}-\frac{1}{1+10 d}\right)\right]=5 \\ & \frac{1}{d}\left[1-\frac{1}{1+10 d}\right]=5 \\ & \frac{1}{d}\left[\frac{1+10 d-1}{1+10 d}\right]=5 \end{aligned}$$

$$\begin{aligned} & \frac{10}{1+10 d}=5 \\ & 1+10 d=2 \\ & d=\frac{1}{10} \\ & 50 d=50 \times \frac{1}{10}=5 \end{aligned}$$

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