Search any question & find its solution
Question:
Answered & Verified by Expert
If the second term in the expansion $\left(\sqrt[13]{\mathrm{a}}+\frac{\mathrm{a}}{\sqrt{\mathrm{a}^{-1}}}\right)^{\mathrm{n}}$ is $14 \mathrm{a}^{5 / 2}$, then $\frac{{ }^{\mathrm{n}} \mathrm{C}_{3}}{{ }^{\mathrm{n}} \mathrm{C}_{2}}=$
Options:
Solution:
1912 Upvotes
Verified Answer
The correct answer is:
4
We have $\mathrm{T}_{2}=14 \mathrm{a}^{\frac{5}{2}}$
$$
\begin{array}{l}
\Rightarrow{ }^{\mathrm{n}} C_{1}\left(\mathrm{a}^{\frac{1}{13}}\right)^{\mathrm{n}-1}\left(\mathrm{a}^{\frac{3}{2}}\right)=14 \mathrm{a}^{\frac{5}{2}} \\
\Rightarrow \mathrm{na}^{\frac{\mathrm{n}-1}{13}+\frac{3}{2}}=14 \mathrm{a}^{\frac{5}{2}} \Rightarrow \mathrm{n}=14 \\
\Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{3}}{{ }^{\mathrm{n}} \mathrm{C}_{2}}=\frac{{ }^{14} \mathrm{C}_{3}}{{ }^{14} \mathrm{C}_{2}}=\frac{12}{3}=4
\end{array}
$$
$$
\begin{array}{l}
\Rightarrow{ }^{\mathrm{n}} C_{1}\left(\mathrm{a}^{\frac{1}{13}}\right)^{\mathrm{n}-1}\left(\mathrm{a}^{\frac{3}{2}}\right)=14 \mathrm{a}^{\frac{5}{2}} \\
\Rightarrow \mathrm{na}^{\frac{\mathrm{n}-1}{13}+\frac{3}{2}}=14 \mathrm{a}^{\frac{5}{2}} \Rightarrow \mathrm{n}=14 \\
\Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{3}}{{ }^{\mathrm{n}} \mathrm{C}_{2}}=\frac{{ }^{14} \mathrm{C}_{3}}{{ }^{14} \mathrm{C}_{2}}=\frac{12}{3}=4
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.