Q62 / 100 · BITSAT 2023 (Memory Based Paper 1) · Mathematics
The middle term in the expansion of $\left(\frac{10}{x}+\frac{x}{10}\right)^{10}$ is
- Correct answerA${ }^{10} \mathrm{C}_5$
- B${ }^{10} \mathrm{C}_6$
- C${ }^{10} \mathrm{C}_5 \frac{1}{x^{10}}$
- D${ }^{10} \mathrm{C}_5 \mathrm{x}^{10}$
Solution
General term $=T_{r+1}$
$={ }^{10} \mathrm{C}_r\left(\frac{10}{x}\right)^{10-r}\left(\frac{x}{10}\right)^r$
Here $n=10$, which is an even number.
Now, $\left[\frac{10}{2}+1\right]^{\text {th }}$ term i.e. $6^{\text {th }}$ term is the middle term.
Hence, middle term $=T_6$
$\mathrm{T}_{5+1}={ }^{10} \mathrm{C}_5\left(\frac{10}{x}\right)^{10-5}\left(\frac{x}{10}\right)^5$
$={ }^{10} \mathrm{C}_5\left(\frac{10}{x}\right)^5\left(\frac{x}{10}\right)^5={ }^{10} \mathrm{C}_5$.
$={ }^{10} \mathrm{C}_r\left(\frac{10}{x}\right)^{10-r}\left(\frac{x}{10}\right)^r$
Here $n=10$, which is an even number.
Now, $\left[\frac{10}{2}+1\right]^{\text {th }}$ term i.e. $6^{\text {th }}$ term is the middle term.
Hence, middle term $=T_6$
$\mathrm{T}_{5+1}={ }^{10} \mathrm{C}_5\left(\frac{10}{x}\right)^{10-5}\left(\frac{x}{10}\right)^5$
$={ }^{10} \mathrm{C}_5\left(\frac{10}{x}\right)^5\left(\frac{x}{10}\right)^5={ }^{10} \mathrm{C}_5$.
Hence, the correct answer is option: 'A'.