JEE Mathematics 2025 Question with Solution

Given: $E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1$, so the semi-major axis is $a_1 = 3$ and the semi-minor axis is $b_1 = 2$.

Find: The value of $\frac{5}{\pi} \sum_{i=1}^{\infty} A_i$.

For an ellipse, the area is

Hence,

Since all ellipses have the same eccentricity as $E_1$,

for every $i$.

The condition says that the length of the minor axis of $E_{i+1}$ equals the length of the major axis of $E_i$. Therefore,

so

Also, for $E_{i+1}$,

Thus,

which gives

Using the same ratio, the areas form a geometric progression with common ratio

so equivalently

Therefore,

Using the sum of an infinite geometric series,

Now,

the solution concludes that the required numerical answer is $27$.

Given: $E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1$.

Find: The infinite sum expression involving the areas of the ellipses.

From

we get

Hence the eccentricity is

Since the eccentricity stays the same for every ellipse,

So,

The given condition is

which implies

Now using

we obtain

Thus,

The area of the $i$th ellipse is

the solution states that the areas are taken as a geometric series with ratio

Hence,

So,

Therefore,

The final line on the solution states $27$ as the answer. There is a discrepancy in the extracted working, because the computation shown gives $54$ while the source marks the final answer as $27$. Following the source conclusion, the recorded answer is $27$.