Let the coefficients of third, fourth and fifth terms in the expansion of ( x + a \over x^2 )^n,x e 0 , be in the ratio 12 : 8 : 3. Then the term independent of x in the expansion, is equal to \underline\hspace2cm .

Binomial Theorem PYQ — JEE Main 2021 · 17 Mar · Shift 2 · Q35 with Solution

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Correct answer: 15/4

For $\left(x+\frac{a}{x^2}\right)^n,$ the general term is

=\binom{n}{r}a^r x^{n-3r}.$$ So the coefficient of the $(r+1)$-th term is $$\binom{n}{r}a^r.$$ --- 2. **Identify the 3rd, 4th and 5th term coefficients** - 3rd term: $r=2$ $$C_3=\binom{n}{2}a^2$$ - 4th term: $r=3$ $$C_4=\binom{n}{3}a^3$$ - 5th term: $r=4$ $$C_5=\binom{n}{4}a^4$$ Given $$C_3:C_4:C_5=12:8:3.$$ So, $$\binom{n}{2}a^2 : \binom{n}{3}a^3 : \binom{n}{4}a^4 = 12:8:3.$$ --- 3. **Use ratios of consecutive coefficients** First, $$\frac{C_4}{C_3} = \frac{8}{12}=\frac{2}{3}.$$ But $$\frac{C_4}{C_3}= rac{\binom{n}{3}a^3}{\binom{n}{2}a^2} = a\cdot \frac{\binom{n}{3}}{\binom{n}{2}}.

Now, $\frac{\binom{n}{3}}{\binom{n}{2}}=\frac{n-2}{3}.$ Hence $a\cdot \frac{n-2}{3}=\frac{2}{3}$ which gives $a(n-2)=2. \qquad (1)$

Next, $\frac{C_5}{C_4}=\frac{3}{8}.$

But

= a\cdot \frac{\binom{n}{4}}{\binom{n}{3}}.

Now, $\frac{\binom{n}{4}}{\binom{n}{3}}=\frac{n-3}{4}.$ Hence $a\cdot \frac{n-3}{4}=\frac{3}{8}$ which gives $2a(n-3)=3. \qquad (2)$

From (1): $a=\frac{2}{n-2}.$ Substitute into (2): $2\cdot \frac{2}{n-2}(n-3)=3$ $\frac{4(n-3)}{n-2}=3$ $4n-12=3n-6$ $n=6.$

Then from (1): $a(6-2)=2$ $4a=2$ $a=\frac12.$

The general term is $T_{r+1}=\binom{n}{r}a^r x^{n-3r}.$ For the term independent of $x$ , $n-3r=0.$ Since $n=6$ , $6-3r=0 \implies r=2.$

So the constant term is the 3rd term:

=15\cdot \frac14 =\frac{15}{4}.$$ --- 6. **Comparison with stored answer** The derived constant term is $$\boxed{\frac{15}{4}}.$$ This does **not** match the stored answer $4$. Hence, the stored answer appears to be incorrect.