Recall the formula for the sum of the first n terms of an arithmetic progression is $S_n=\frac{n}{2}(2a+(n-1)d)$.

The sum of the first $\var{n1}$ terms of an arithmetic progression is $\var{s1}$

$\frac{\var{n1}}{2}(2a+\simplify{{n1}-1}d)=\var{s1}$                               

$\var{n1}a+\simplify{{n1}*({n1}-1)/2}d=\var{s1}$                                       equation (i)

The sum of the first $\var{n2}$ terms of an arithmetic progression is $\var{s2}$

$\frac{\var{n2}}{2}(2a+\simplify{{n2}-1}d)=\var{s2}$                               

$\var{n2}a+\simplify{{n2}*({n2}-1)/2}d=\var{s2}$                                      equation (ii)

We can eliminate the $a$ term.

$\simplify{{n2}*{n1}}a+\simplify{{n2}*{n1}*({n1}-1)/2}d=\simplify{{n2}*{s1}}$                               equation (i) * $\var{n2}$

$\simplify{{n2}*{n1}}a+\simplify{{n1}*{n2}*({n2}-1)/2}d=\simplify{{n1}*{s2}}$                               equation (ii) * $\var{n1}$

Subtracting gives

$\simplify{{n2}*{n1}*({n1}-{n2})/2}d=\simplify{{n2}*{s1}-{n1}*{s2}}$

$d=\frac{\simplify{{n2}*{s1}-{n1}*{s2}}}{\simplify{{n2}*{n1}*({n1}-{n2})/2}}$

$d=\simplify{2*({n2}*{s1}-{n1}*{s2})/({n2}*{n1}*({n1}-{n2}))}$

Inserting this value in for $d$ in equation (i) gives

$\var{n1}a+\simplify{{n1}*({n1}-1)/2}(\simplify{2*({n2}*{s1}-{n1}*{s2})/({n2}*{n1}*({n1}-{n2}))})=\var{s1}$

$\var{n1}a+(\simplify{({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2}))})=\var{s1}$

$\var{n1}a=\var{s1}-(\simplify{({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2}))})$

$\var{n1}a=\simplify{{s1}-({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2}))}$

$a=\simplify{({s1}-({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2})))/{n1}}$

$a=\var{a}$