Q80 / 120 · VITEEE 2019 · Mathematics

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 $\frac{{ }^{\mathrm{n}} \mathrm{C}_{3}}{{ }^{\mathrm{n}} \mathrm{C}_{2}}=$

  • Correct answer
    A
    4
  • B
    3
  • C
    12
  • D
    6

Solution

We have $\mathrm{T}_{2}=14 \mathrm{a}^{\frac{5}{2}}$ $$ \begin{array}{l} \Rightarrow{ }^{\mathrm{n}} C_{1}\left(\mathrm{a}^{\frac{1}{13}}\right)^{\mathrm{n}-1}\left(\mathrm{a}^{\frac{3}{2}}\right)=14 \mathrm{a}^{\frac{5}{2}} \\ \Rightarrow \mathrm{na}^{\frac{\mathrm{n}-1}{13}+\frac{3}{2}}=14 \mathrm{a}^{\frac{5}{2}} \Rightarrow \mathrm{n}=14 \\ \Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{3}}{{ }^{\mathrm{n}} \mathrm{C}_{2}}=\frac{{ }^{14} \mathrm{C}_{3}}{{ }^{14} \mathrm{C}_{2}}=\frac{12}{3}=4 \end{array} $$

Hence, the correct answer is option: 'A'.

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