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The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1 + \alpha x)^4$ and of $(1 - \alpha x)^6$ is the same if $\alpha$ equals
$-\frac{5}{3}$
$\frac{10}{3}$
$-\frac{3}{10}$
$\frac{3}{5}$
Solution
The middle term in the expansion of $(1+\alpha x )^{ n }$ is $\left(\frac{ n }{2}+1\right)^{\text {th }}$ term.
General term in the expansion of $(1+\alpha x )^4$ is $4_{ C _{ r }} a _{ r } x ^{ r }$
General term in the expansion of $(1-\alpha x)^6$ is $6_{C_1 r} a^r x^r$
The middle term in the expansion of $(1+\alpha x )^4$ is $\left(\frac{4}{2}+1\right)^{\text {th }}=3^{\text {rd }}$ term.
The middle term in the expansion of $(1-\alpha x )^6$ is $\left(\frac{6}{2}+1\right)^{\text {th }}=4^{\text {th }}$ term.
Given: $4_{ C _2 \alpha^2}=-6_{ C _3 \alpha^3}$
$\Rightarrow \alpha =\frac{-4_{ C 2}}{6_{ C 3}}$
$=\frac{-\frac{4 \times 3}{2 \times 1}}{\frac{6 \times 5 \times 4}{3 \times 2 \times 1}}$
$=\frac{-6}{20}=\frac{-3}{10}$