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 ________.

General term in the expansion of

$$\left(x+\frac{a}{x^2}\right)^n$$

is

$$T_{r+1}=\binom nr x^{n-r}\left(\frac{a}{x^2}\right)^r$$

$$=\binom nr a^r x^{n-3r}$$

Hence coefficients of third, fourth and fifth terms are:

$$T_3:\binom n2 a^2$$

$$T_4:\binom n3 a^3$$

$$T_5:\binom n4 a^4$$

Given,

$$\binom n2 a^2:\binom n3 a^3:\binom n4 a^4=12:8:3$$

Taking ratios,

$$\frac{\binom n2 a^2}{\binom n3 a^3}=\frac{12}{8}=\frac32$$

$$\frac{\frac{n(n-1)}2}{\frac{n(n-1)(n-2)}6}\cdot\frac1a=\frac32$$

$$\frac{3}{n-2}\cdot\frac1a=\frac32$$

$$a(n-2)=2\quad\cdots(1)$$

Also,

$$\frac{\binom n3 a^3}{\binom n4 a^4}=\frac83$$

$$\frac{\frac{n(n-1)(n-2)}6}{\frac{n(n-1)(n-2)(n-3)}{24}}\cdot\frac1a=\frac83$$

$$\frac{4}{n-3}\cdot\frac1a=\frac83$$

$$a(n-3)=\frac32\quad\cdots(2)$$

From (1) and (2),

$$\frac2{n-2}=\frac{3/2}{n-3}$$

$$4(n-3)=3(n-2)$$

$$4n-12=3n-6$$

$$n=6$$

Then from

$$a(n-2)=2,$$

$$4a=2$$

$$a=\frac12$$

Now term independent of $$x$$ occurs when

$$n-3r=0$$

$$6-3r=0$$

$$r=2$$

Hence required term is

$$T_3=\binom62\left(\frac12\right)^2$$

$$=15\cdot\frac14$$

$$=\frac{15}{4}$$

Therefore, the term independent of $$x$$ is

$$\boxed{\frac{15}{4}}$$