Q113 / 170 · Manipal (MET) 2018 · Mathematics
The first three terms in the expansion of $(1+a x)^n(n \neq 0)$ are 1, $6 x$ and $16 x^2$. Then, the values of $a$ and $n$ are respectively
- A2 and 9
- Correct answerB$\frac{2}{3}$ and 9
- C3 and 2
- D$\frac{3}{2}$ and 6
Solution
$\because(1+a x)^n=1+a x n+\frac{n(n-1)}{2!}(a x)^2+\ldots$
Given that,
$\begin{aligned}
& a x n & =6 x \\
\Rightarrow & a n & =6...(i)
\end{aligned}$
and $\quad \frac{a^2 n(n-1)}{2}=16$...(ii)
On solving Eqs. (i) and (ii), we get $a=\frac{2}{3}$ and $n=9$
Hence, the correct answer is option: 'B'.