Mathematics · Binomial Theorem

JEE Advanced 2025 — Paper 2 — Question 26

Let $a_{0}, a_{1}, \ldots, a_{23}$ be real numbers such that $\left(1+\frac{2}{5} x\right)^{23}=\sum_{i=0}^{23} a_{i} x^{i}$ for every real number $x$ . let $a_{r}$ be the largest among the numbers $a_{j}$ for $0 \leq j \leq 23$ . The the value of $r$ is $\qquad$

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

For $\mathrm{x}=1$

$\left(1+\frac{2}{5}\right)^{23}=a_{0}+a_{1}+a_{2}+\ldots+a_{23}$

for numerically greatest term

$\frac{n+1}{1+\left|\frac{a}{b}\right|}=\frac{23+1}{1+\frac{5}{2}}=\frac{48}{7}$

$\Rightarrow\left[\frac{48}{7}\right]=6=\mathrm{m}$ (where [.] greatest integer function)

so, $\mathrm{T}_{7}$ is numerical greatest term\ Hence $\mathrm{r}=6$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 2
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Applications of Binomial Theorem