Let S = N cup\ 0\ . Define a relation R from S to R by : R = \ ( x , y ): log e y = x log e (2/5 ), x in S , y in R \ Then, the sum of all the elements in the range of R is equal to

Mathematics · Sets and Relations

JEE Main 2025 — 29 January, Evening Shift — Question 48

Let $\mathrm{S}=\mathbf{N} \cup\{0\}$ . Define a relation $\mathbf{R}$ from $\mathbf{S}$ to $\mathbf{R}$ by :

$\mathbf{R}=\left\{(\mathrm{x}, \mathrm{y}): \log _{\mathrm{e}} \mathrm{y}=\mathrm{x} \log _{\mathrm{e}}\left(\frac{2}{5}\right), \mathrm{x} \in \mathrm{~S}, \mathrm{y} \in \mathbf{R}\right\}$ Then, the sum of all the elements in the range of $\mathbf{R}$ is equal to

Option A:
$\frac{3}{2}$
Option B:
$\frac{5}{3}$
Correct
Option C:
$\frac{10}{9}$
Option D:
$\frac{5}{2}$

Answer: B

Step-by-step solution

$S=\{0,1,2,3 \ldots\}$ .

$\log _{\mathrm{e}} \mathrm{y}=x \log _{\mathrm{e}}\left(\frac{2}{5}\right)$

$\Rightarrow \mathrm{y}=\left(\frac{2}{5}\right)^{\mathrm{x}}$

Required $\operatorname{Sum}=1+\left(\frac{2}{5}\right)^{1}+\left(\frac{2}{5}\right)^{2}+\left(\frac{2}{5}\right)^{3}+\ldots . .-=\frac{1}{1-\frac{2}{5}}=\frac{5}{3}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Mathematics
Chapter: Sets and Relations
Topic: Relations