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JEE Advanced 2025Paper 2

Exam: JEE Advanced
Year: 2025
Paper: Paper 2
Subject: Mathematics
Chapter: Binomial Theorem

JEE Advanced 2025 · Paper 2 · Mathematics · Binomial Theorem

Let a_0, a_1, , a_ 23 be real numbers such that (1+(2)/(5) x)^ 23...

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 ______

Correct answer

0

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

r=6

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