A) \[{{T}_{4}}\]
B) \[{{T}_{3}}\]
C) \[{{T}_{5}}\]
D) None of these
Correct Answer: B
Solution :
\[{{(4-3x)}^{7}}={{4}^{7}}{{\left( 1-\frac{3x}{4} \right)}^{2}}\] Consider, \[{{\left( 1-\frac{3x}{4} \right)}^{7}}\] \[\frac{{{T}_{r+1}}}{{{T}_{r}}}=\frac{7-r+1}{r}.\frac{3}{4}.\frac{2}{3}\ge 1\] \[\Rightarrow \] \[r\le 2.66\] \[\therefore \] \[r=2\] Hence, the greatest term is \[{{T}_{3}}\].
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