A) 2 and 9
B) 3 and 2
C) 2/3 and 9
D) 3/2 and 6
Correct Answer: C
Solution :
[c] : We have, \[{{T}_{1}}{{=}^{n}}{{C}_{0}}=1\] ...(i) \[{{T}_{2}}{{=}^{n}}{{C}_{1}}ax=6x\] ...(ii) \[{{T}_{3}}{{=}^{n}}{{C}_{2}}{{(ax)}^{2}}=16{{x}^{2}}\] ...(iii) From (ii), \[\frac{n!}{(n-1)!}a=6\Rightarrow na=6\] ...(iv) From (iii),\[\frac{n(n-1)}{2}{{a}^{2}}=16\] ...(v) Solving (iv) and (v), we have \[a=\frac{2}{3}\]and n = 9.
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