1. We need the coefficient of the middle term in the expansions of:

$(1+\alpha x)^4 \quad \text{and} \quad (1-\alpha x)^6$

and these coefficients are given to be equal.

2. Middle term of $(1+\alpha x)^4$

For $(a+b)^n$, the number of terms is $n+1$.

Here, $n=4$, so number of terms $=5$, hence there is a unique middle term: the $3^{\text{rd}}$ term.

General term is

$T_{r+1}=\binom{4}{r}(1)^{4-r}(\alpha x)^r$

Middle term corresponds to $r=2$:

$T_3=\binom{4}{2}(\alpha x)^2=6\alpha^2 x^2$

So, the coefficient of the middle term is

$6\alpha^2$

3. Middle term of $(1-\alpha x)^6$

Here, $n=6$, so number of terms $=7$, hence the middle term is the $4^{\text{th}}$ term.

General term is

$T_{r+1}=\binom{6}{r}(1)^{6-r}(-\alpha x)^r$

Middle term corresponds to $r=3$:

$T_4=\binom{6}{3}(-\alpha x)^3=20(-\alpha^3)x^3=-20\alpha^3 x^3$

So, the coefficient of the middle term is

$-20\alpha^3$

4. Equate the coefficients

Given that the coefficients are the same:

$6\alpha^2=-20\alpha^3$

Bring all terms to one side:

$6\alpha^2+20\alpha^3=0$

Factor:

$2\alpha^2(3+10\alpha)=0$

So,

$\alpha^2=0 \quad \text{or} \quad 3+10\alpha=0$

Hence,

$\alpha=0 \quad \text{or} \quad \alpha=-\frac{3}{10}$

Since $0$ is not among the options, the required value is

$\boxed{-\frac{3}{10}}$

5. Check options

• A: $\frac{3}{5}$ ❌
• B: $\frac{10}{3}$ ❌
• C: $-\frac{3}{10}$ ✅
• D: $-\frac{5}{3}$ ❌

So the correct option is C.