JEE Advance - Mathematics (2025 - Paper 2 Online - No. 10)
Let $a_0, a_1, \ldots, a_{23}$ be real numbers such that
$$ \left(1+\frac{2}{5} x\right)^{23}=\sum\limits_{i=0}^{23} a_i x^i $$
for every real number $x$. Let $a_r$ be the largest among the numbers $a_j$ for $0 \leq j \leq 23$. Then the value of $r$ is ____________.
Answer
6
Explanation
For $\mathrm{x}=1$
$$ \left(1+\frac{2}{5}\right)^{23}=a_0+a_1+a_2+\ldots+a_{23} $$
for numerically greatest term
$$ \begin{aligned} & \frac{\mathrm{n}+1}{1+\left|\frac{\mathrm{a}}{\mathrm{~b}}\right|}=\frac{23+1}{1+\frac{5}{2}}=\frac{48}{7} \\ & \Rightarrow\left[\frac{48}{7}\right]=6=\mathrm{m} \quad \text { (where [.] greatest integer function) } \end{aligned} $$
so, $\mathrm{T}_7$ is numerical greatest term
Hence $\mathrm{r}=6$
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