JEE Main 2025 Jan 29 Shift 2, Mathematics Q7: Summation of Series

JEE Main2025Jan 29, Shift 2Mathematics

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

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

Solution

Exponentiating, $y = (\frac{2}{5})^{x}$ for $x \in S = {0, 1, 2, 3, \dots}$ .

The range is the geometric progression ${1, \frac{2}{5}, (\frac{2}{5})^{2}, (\frac{2}{5})^{3}, \dots}$ with first term $1$ and common ratio $\frac{2}{5}$ .

Sum $= \frac{1}{1 - \frac{2}{5}} = \frac{1}{3/5} = \frac{5}{3}$ .

Concept behind this question

Summation of SeriesNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 29 Shift 2, Mathematics Q7: Summation of Series

Concept behind this question

More previous year questions on Summation of Series

Practice more Mathematics