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(C) The general term is $T_{r+1} = {}^{n}C_{r} x^{n-r} (\frac{a}{x^2})^r = {}^{n}C_{r} a^r x^{n-3r}$.The coefficients of the third,fourth,and fifth te

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$4$

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(C) The general term is $T_{r+1} = {}^{n}C_{r} x^{n-r} (\frac{a}{x^2})^r = {}^{n}C_{r} a^r x^{n-3r}$.The coefficients of the third,fourth,and fifth terms are ${}^{n}C_{2} a^2$,${}^{n}C_{3} a^3$,and ${}^{n}C_{4} a^4$ respectively.Given the ratio ${}^{n}C_{2} a^2 : {}^{n}C_{3} a^3 : {}^{n}C_{4} a^4 = 12 : 8 : 3$.From $\frac{{}^{n}C_{2} a^2}{{}^{n}C_{3} a^3} = \frac{12}{8} = \frac{3}{2}$,we get $\frac{3}{a(n-2)} = \frac{3}{2} \implies a(n-2) = 2$.From $\frac{{}^{n}C_{3} a^3}{{}^{n}C_{4} a^4} = \frac{8}{3}$,we get $\frac{4}{a(n-3)} = \frac{8}{3} \implies a(n-3) = \frac{3}{2}$.Solving these,we find $n=6$ and $a=\frac{1}{2}$.The term independent of $x$ occurs when $n-3r = 0$,so $6-3r = 0 \implies r=2$.The term is ${}^{6}C_{2} a^2 = 15 	imes (\frac{1}{2})^2 = \frac{15}{4} = 3.75$.The nearest integer is $4$.

Let the coefficients of the third,fourth,and fifth terms in the expansion of $(x + \frac{a}{x^2})^n, x \neq 0,$ be in the ratio $12 : 8 : 3$. Then the term independent of $x$ in the expansion is equal to ...... .

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