We have

$a_2-a_1=a_3-a_2=\mathrm{...............}{a}_{2k}-a_{2k-1}=d$

Hence,$a_1^2-a_2^2=(a_1-a_2)(a_1+a_2)=-d(a_1+a_2)$   $a_3^2-a_4^2=(a_3-a_4)(a_3+a_4)=-d(a_3+a_4)$

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$a_{2k-1}^2-a_{2k}^2=(a_{2k-1}-a_{2k})(a_{2k-1}+a_{2k})=-d\left(a_{2k-1}+a_{2k}\right)$

$$\begin{gathered} \therefore a_1^2-a_2^2+a_3^2-a_4^2+\mathrm{............}+a_{2k-1}^2-a_{2k}^2=-d\left(a_1+a_2+a_3+.......+a_{2k}\right) \\ =-d.\frac{2k}{2}(a_1+a_{2k})=-dk(a_1+a_{2k}) \end{gathered}$$

$but{a}_{2k}=a_1+\left(2k-1\right)d\Rightarrow -d=\frac{a_1-a_{2k}}{2k-1}$

$\therefore$  the required sum =$\frac{k}{2k-1}\left(a_1{ }^2-a_{2k}{ }^2\right)\Rightarrow M=\frac{k}{2k-1}$