The coefficients of the middle terms in the binomial expansion in powers of $x$  of ${\left(1+ax\right)}^4$  and of ${\left(1-ax\right)}^6$  are same if $a$  equals

Given  expansion ${\left(1+ax\right)}^4$ is,

We have general term in the expansion ${\left(x+a\right)}^n$

$(\therefore T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$ be the expansion of ${(x+a)}^n)$

The middle term is ${\left(\frac{4}{2}+1\right)}^{th}$  i.e. $3^{rd}$  term

$3^{rd}$  term :

$T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$

$T_3=T_{2+1}={}^4{C}_2{\left(1\right)}^2{\left(ax\right)}^2$

$T_3=T_{2+1}{=}^4{C}_2{a}^2{x}^2$

And in ${\left(1-ax\right)}^6,{\left(\frac{6}{2}+1\right)}^{th}$  i.e. $4^{th}$  term is the middle term.

$4^{th}$  term :

$T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$

And in ${\left(1-ax\right)}^6,T_4=T_{3+1}={}^6{C}_3{\left(1\right)}^3{\left(-ax\right)}^3$

According to question, ${}^4{C}_2{a}^2={}^6{C}_3{\left(-a\right)}^3$ $\begin{array}{c}({\therefore }^n{C}_r=\frac{n!}{(n-r)!r!})\\ \end{array}$

$\Rightarrow 6a^2=-20a^3$  $\Rightarrow a=-\frac{6}{20}=-\frac{3}{10}\left[a\ne 0\right]$