JEE Advanced 2025 · Paper 2 · Mathematics Q10

MathematicsNumerical+4Question 10 of 48

Let $a_0, a_1, \ldots, a_{23}$ be real numbers such that

$\left(1 + \frac25x\right)^{23} = \sum_{i=0}^{23}a_ix^i$

for every real number $x$ . Let $a_r$ be the largest among the numbers $a_j$ for $0 \le j \le 23$ . Then the value of $r$ is _____.

Your answer

Answer and worked solution

Official answer: 6

$a_r={}^{23}C_r\left(\frac25\right)^r$ , so
$\frac{a_{r+1}}{a_r}=\frac{23-r}{r+1}\cdot\frac25.$

This ratio exceeds 1 exactly when $46-2r>5r+5$ , i.e. $r<\frac{41}7\approx5.86$ .

So the coefficients increase up to $a_6$ and then decrease. The largest is $a_6$ , so $r=6$ .

JEE Advanced 2025 · Paper 2, Mathematics question 10