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

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Sets Relations and Functions

Let

S = N \cup {0}

. Define a relation

R

from S to

R

by :

R = {(x, y) : lo g_{e} y = x lo g_{e} (\frac{2}{5}), x \in S, y \in R}

.
Then, the sum of all the elements in the range of

R

is equal to

Answer: (B)

lo g_{e} y = x lo g_{e} (\frac{2}{5}) \Rightarrow y = (\frac{2}{5})^{x}

, and

x

ranges over

S = {0, 1, 2, \dots}

.

The range is the infinite geometric progression

1, \frac{2}{5}, (\frac{2}{5})^{2}, \dots

with first term

1

and common ratio

\frac{2}{5}

sum = \frac{1}{1 - \frac{2}{5}} = \frac{5}{3}

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option