Let $S=\mathbf{N} \cup\{0\}$. Define a relation $R$ from $S$ to $\mathbf{R}$ by : $\mathrm{R}=\left\{(x, y):…
$\mathrm{R}=\left\{(x, y): \log _{\mathrm{e}} y=x \log _{\mathrm{e}}\left(\frac{2}{5}\right), x \in \mathrm{~S}, y \in \mathbf{R}\right\}$
Then, the sum of all the elements in the range of $R$ is equal to :
- $\frac{10}{9}$
- $\frac{3}{2}$
- $\frac{5}{2}$
- $\frac{5}{3}$
Solution

Required
$\text { Sum }=1+\left(\frac{2}{5}\right)^1+\left(\frac{2}{5}\right)^2+\left(\frac{2}{5}\right)^3+\ldots . .-=\frac{1}{1-\frac{2}{5}}=\frac{5}{3}$ *
Asked in: JEE Main 2025 (29 Jan Shift 2)