Let S = N 0. Define a relation R from S to R by: R = (x, y): e y = x e ( 25), x S, y R. Then, the sum of all the elements in the range of R is equal to:

Given: S = N 0 and R = (x, y): e y = x e ( 25), x S, y R. Find: The sum of all elements in the range of R. From the given relation, e y = x e ( 25) Exponentiating both sides, y = ( 25)^x Since x S = N 0, the range values are 1, ( 25), ( 25)^2, ( 25)^3, So their sum is the infinite geometric series 1 + ( 25) + ( 25)^2 + ( 25)^3 + Here, the first term is a = 1 and the common ratio is r = 25. Since |r | < 1, the infinite series converges. Using the formula Sum = a1-r we get Sum = 11 - 25 = 1 35 = 53 Therefore, the sum of all the elements in the range of R is 53. The correct option is B.

Let S = N 0. Define a relation R from S to R by: R = (x, y): e y… — Solution | JEE Mathematics

MCQEasyJEE 2025Sum of Series

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

R = \left\{ (x, y) : \log_e y = x \log_e \left(\frac{2}{5}\right), x \in S, y \in \mathbb{R} \right\}.

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
$\frac{5}{2}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $S = \mathbb{N} \cup \{0\}$ and

R = \left\{ (x, y) : \log_e y = x \log_e \left(\frac{2}{5}\right), x \in S, y \in \mathbb{R} \right\}.

Find: The sum of all elements in the range of $R$ .

From the given relation,

\log_e y = x \log_e \left(\frac{2}{5}\right)

Exponentiating both sides,

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

Since $x \in S = \mathbb{N} \cup \{0\}$ , the range values are

1, \left(\frac{2}{5}\right), \left(\frac{2}{5}\right)^2, \left(\frac{2}{5}\right)^3, \ldots

So their sum is the infinite geometric series

1 + \left(\frac{2}{5}\right) + \left(\frac{2}{5}\right)^2 + \left(\frac{2}{5}\right)^3 + \cdots

Here, the first term is $a = 1$ and the common ratio is $r = \frac{2}{5}$ . Since $\left|r\right| < 1$ , the infinite series converges. Using the formula

\text{Sum} = \frac{a}{1-r}

we get

\text{Sum} = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3}

Therefore, the sum of all the elements in the range of $R$ is $\frac{5}{3}$ . The correct option is B.

Common mistakes

Taking $S = \mathbb{N}$ instead of $S = \mathbb{N} \cup \{0\}$ is incorrect because it misses the term corresponding to $x = 0$ . That first term is $1$ , so omitting it changes the entire sum. Always list the starting value of $x$ carefully before writing the range.
Not exponentiating the logarithmic equation correctly is a conceptual error. From
$\log_e y = x \log_e \left(\frac{2}{5}\right)$
you must use log properties to obtain
$y = \left(\frac{2}{5}\right)^x$
not $y = x\left(\frac{2}{5}\right)$ or any linear expression.
Using the finite geometric progression formula instead of the infinite sum formula is wrong here because the range contains infinitely many values. Since $\left|\frac{2}{5}\right| < 1$ , use
$\text{Sum} = \frac{a}{1-r}$
for the infinite geometric series.

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