Q172 / 270 · NDA 2019 (Phase 1) · Mathematics

In the expansion of $(1+a x)^{\mathrm{n}}$, the first three terms are respectively $1,12 x$ and $64 x^2$. What is n equal to?

  • A
    6
  • Correct answer
    B
    9
  • C
    10
  • D
    12

Solution

The first three terms in expansion of $(1+\mathrm{ax})^{\mathrm{n}}$ are ${ }^n C_0,{ }^n C_1 a x,{ }^n C_2 a^2 x^2$ Given, ${ }^n C_0=1 ;{ }^n C_1 a x=12 x ;{ }^n C_2 a^2 x^2=64 x^2$ $\begin{aligned} & \Rightarrow \mathrm{nax}=12 \mathrm{x} ; \frac{n(n-1)}{2} a^2=64 \\ & \Rightarrow \mathrm{na}=12 \Rightarrow a=\frac{12}{n} \\ & \therefore \frac{n(n-1)}{2} a^2=64 \Rightarrow \frac{n(n-1)}{2} \times \frac{144}{n^2}=64 \\ & \Rightarrow \frac{n-1}{n}=\frac{64 \times 2}{144}=\frac{8}{9} \\ & \therefore \mathrm{n}=9\end{aligned}$

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

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