The first 3 terms in the expansion of (1+ax)n (n ≠ 0) are 1, 6x and 16x

Question

The first 3 terms in the expansion of $(1 + a x)^{n}$ (n ≠ 0) are 1, 6x and 16 $x^{2}$ . Then the value of a and n are respectively

Options:

A . 2 and 9

B . 3 and 2

C .

\frac{2}{3}

and 9

D .

\frac{3}{2}

and 6

Answer: Option C
:
C

$n_{c_{1}} . a x = 6 x; n_{c_{2}} (a x)^{2} = 16 x^{2} \Rightarrow a n = 6; \frac{n (n - 1)}{2} a^{2} = 16 \Rightarrow a = \frac{2}{3} \Rightarrow n = 9$

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