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7.Binomial Theorem
medium
Let the coefficients of third, fourth and fifth terms in the expansion of $\left(x+\frac{a}{x^{2}}\right)^{n}, x \neq 0,$ be in the ratio $12: 8: 3 .$ Then the term independent of $x$ in the expansion, is equal to ...... .
A
$5$
B
$3$
C
$4$
D
$6$
(JEE MAIN-2021)
Solution
$T _{ r +1} ={ }^{ n } C _{ r }( x )^{ n – r }\left(\frac{ a }{ x ^{2}}\right)^{ r }$
$={ }^{n} C _{ r } a ^{ r } x ^{ n -3 r }$
${ }^{ n } C _{2} a ^{2}:{ }^{ n } C _{3} a ^{3}:{ }^{ n } C _{4} a ^{4}=12: 8: 3$
After solving
$n =6, a =\frac{1}{2}$
For term independent of $x ^{\prime} \Rightarrow n =3 r$
$r =2$
$\therefore$ Coefficient is ${ }^{6} C _{2}\left(\frac{1}{2}\right)^{2}=\frac{15}{4}$
Nearest integer is $4 .$
Standard 11
Mathematics
