JEE Main 2024 — 9 April, Shift 1 — Question 12

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

$\frac{1}{\mathrm{~d}}\left[\frac{(1+\mathrm{d})-1}{1 \cdot(1+\mathrm{d})}+\frac{(1+2 \mathrm{~d})-(1-\mathrm{d})}{(1+\mathrm{d})(1+2 \mathrm{~d})}\right]+$ $\qquad$

$\frac{(1+10 d)-(1+9 d)}{(1+9 d)(1+10 d)}=5$

$\frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}\right)+\right.$

$\left.\left(\frac{1}{1+9 d}-\frac{1}{1+10 d}\right)\right]=5$

$\frac{1}{d}\left[1-\frac{1}{(1+10 \mathrm{~d})}\right]=5$

$\frac{10 \mathrm{~d}}{1+10 \mathrm{~d}}=5 \mathrm{~d}$

$50 \mathrm{~d}=5$