A) 2 and 9
B) 3 and 2
C) \[\frac{2}{3}\]and 9
D) \[\frac{3}{2}\]and 6
E) \[\frac{-2}{3}\] and 9
Correct Answer: C
Solution :
\[\because \] \[{{(1+ax)}^{n}}=1+axn+\frac{n(n-1)}{2!}{{(ax)}^{2}}+....\] Given that \[a\,x\,n=6x\] \[\Rightarrow \] \[an=6\] ...(i) and \[\frac{{{a}^{2}}n(n-1)}{2}=16\] ?..(ii) On solving Eqs. (i) and (ii), we get \[a=\frac{2}{3}\]and\[n=9\]You need to login to perform this action.
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