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 ____.

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Binomial Theorem

Let

a_{0}, a_{1}, \dots ., a_{23}

be real numbers such that

(1 + \frac{2}{5} x)^{23} = i = 0 \sum 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 ____.

Answer: 6

For the binomial expansion

(a + b x)^{n}

, the numerically greatest term occurs at position

m

where:

m = ⌊ \frac{n + 1}{1 + ∣ a / b ∣} ⌋

Here

a = 1

b = \frac{2}{5}

n = 23

m = ⌊ \frac{23 + 1}{1 + \frac{5}{2}} ⌋ = ⌊ \frac{24}{7/2} ⌋ = ⌊ \frac{48}{7} ⌋ = ⌊ 6.857 ⌋ = 6

The

(m + 1)

-th term is

T_{7}

, which corresponds to

a_{6}

.

Therefore

r = 6

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer