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7.Binomial Theorem
medium
The first $3$ terms in the expansion of ${(1 + ax)^n}$ $(n \ne 0)$ are $1, 6x$ and $16x^2$. Then the value of $a$ and $n$ are respectively
A
$2$ and $9$
B
$3$ and $2$
C
$2/3$ and $9$
D
$3/2$ and $6$
Solution
(c) ${T_1} = {}^n{C_0} = 1$ …..$(i)$
${T_2} = {}^n{C_1}ax = 6x$ …..$(ii)$
${T_3} = {}^n{C_2}{(ax)^2} = 16{x^2}$ …..$(iii)$
From $(ii)$, $\frac{{n!}}{{\left( {n – 1} \right)\,!}}a = 6$
==> $n\,a\, = \,6$ …..$(iv)$
From $(iii)$, $\frac{{n(n – 1)}}{2}\,{a^2} = 16$ ….$.(v)$
Only $(c)$ is satisfying equation $(iv)$ and $(v)$.
Standard 11
Mathematics
