Question:
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
Solution:
$\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{\mathrm{n}-\mathrm{r}}\left(\frac{\mathrm{a}}{\mathrm{x}^{2}}\right)^{\mathrm{r}}$
$={ }^{n} C_{r} a^{r} x^{n-3 r}$
${ }^{n} C_{2} a^{2}:{ }^{n} C_{3} a^{3}:{ }^{n} C_{4} a^{4}=12: 8: 3$
After solving
$\mathrm{n}=6, \mathrm{a}=\frac{1}{2}$
For term independent of ' $x$ ' $\Rightarrow n=3 r$
$r=2$
$\therefore$ Coefficient is ${ }^{6} \mathrm{C}_{2}\left(\frac{1}{2}\right)^{2}=\frac{15}{4}$
Nearest integer is 4 .