JEE MAIN - Mathematics (2025 - 29th January Evening Shift - No. 1)
Let $\mathrm{S}=\mathbf{N} \cup\{0\}$. Define a relation R from S to $\mathbf{R}$ by :
$$ \mathrm{R}=\left\{(x, y): \log _{\mathrm{e}} y=x \log _{\mathrm{e}}\left(\frac{2}{5}\right), x \in \mathrm{~S}, y \in \mathbf{R}\right\} . $$
Then, the sum of all the elements in the range of $R$ is equal to :
$\frac{3}{2}$
$\frac{10}{9}$
$\frac{5}{2}$
$\frac{5}{3}$
Explanation
$$\begin{aligned} & S=\{0,1,2,3 \ldots . .\} \\ & \log _{\mathrm{e}} \mathrm{y}=\log _{\mathrm{e}}\left(\frac{2}{5}\right) \\ & \Rightarrow \mathrm{y}=\left(\frac{2}{5}\right)^{\mathrm{x}} \end{aligned}$$
_29th_January_Evening_Shift_en_1_1.png)
Required
$$\text { Sum }=1+\left(\frac{2}{5}\right)^1+\left(\frac{2}{5}\right)^2+\left(\frac{2}{5}\right)^3+\ldots . .-=\frac{1}{1-\frac{2}{5}}=\frac{5}{3}$$
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