A) 11
B) 10
C) 9
D) 12
Correct Answer: A
Solution :
rth term in the expansion of\[{{\left( 3x-\frac{2}{{{x}^{2}}} \right)}^{15}}\]is \[{{T}_{r}}{{=}^{15}}{{C}_{r-1}}{{(3x)}^{15-r+1}}{{\left( -\frac{2}{{{x}^{2}}} \right)}^{r-1}}\] \[{{=}^{15}}{{C}_{r-1}}{{(3)}^{15-r+1}}{{(-2)}^{r-1}}{{(x)}^{15-3r+3}}\] For the term independent of \[x,\] put \[15-3r+3=0\Rightarrow r=6\] and rth term from the end\[=(15-6+2)=11\]
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