JEE Main 2025 Jan 28 Shift 1, Mathematics Q25: Summation of Series

JEE Main2025Jan 28, Shift 1Mathematics

Let $E_{1} : \frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ be an ellipse. Ellipses $E_{i}$ 's are constructed such that their centres and eccentricities are same as that of $E_{1}$ , and the length of minor axis of $E_{i}$ is the length of major axis of $E_{i + 1}$ ( $i \geq 1$ ). If $A_{i}$ is the area of the ellipse $E_{i}$ , then $\frac{5}{π} (i = 1 \sum \infty A_{i})$ is equal to _____.

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Solution

For $E_{1}$ , $a_{1} = 3$ , $b_{1} = 2$ , so eccentricity $e = 1 - \frac{4}{9} = \frac{5}{3}$ .

Each subsequent ellipse $E_{i + 1}$ has the same eccentricity as $E_{i}$ , and its major axis equals the minor axis of $E_{i}$ . In other words, the major semi-axis shrinks by the factor $\frac{b _{i}}{a _{i}} = 1 - e^{2} = \frac{2}{3}$ , and because eccentricity is preserved, all linear dimensions of the next ellipse scale by $\frac{2}{3}$ .

The area of an ellipse is $π ab$ , which scales by $(\frac{2}{3})^{2} = \frac{4}{9}$ between consecutive ellipses.

$A_{1} = π (3) (2) = 6 π$ , and the sequence ${A_{i}}$ is geometric with ratio $\frac{4}{9}$ .

$\sum_{i = 1}^{\infty} A_{i} = \frac{6 π}{1 - 4/9} = \frac{6 π}{5/9} = \frac{54 π}{5}$ .

$\frac{5}{π} \cdot \frac{54 π}{5} = 54$ .

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JEE Main 2025 Jan 28 Shift 1, Mathematics Q25: Summation of Series

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