Find terms number of arithmetic progression.

I had an exam today, within the exam, this question was the hardest.

If we have a arithmetic progression, its number of terms is $even$, total of it's $even$ terms = $30$, total of it's $odd$ terms = $24$.

the difference between the last term and the first one = $10.5$

(If nothing clear, sorry for it, I tried to translate the question into english)

Answer

Let $a,d,2m$ be the first term, the common difference, the number of terms respectively where $m\in\mathbb N$.

This answer supposes that "total of it's even terms $=30$" means that
$$(a+d)+(a+3d)+\cdots +(a+(2m-1)d)=\sum_{i=1}^{m}(a+(2i-1)d)=30,$$
i.e.
$$am+2d\cdot\frac{m(m+1)}{2}-dm=30\tag1$$

Also, this answer supposes that "total of it's odd terms $=24$" means that
$$a+(a+2d)+\cdots +(a+(2m-2)d)=\sum_{i=1}^{m}(a+(2i-2)d)=24,$$
i.e.
$$am+2d\cdot\frac{m(m+1)}{2}-2dm=24\tag2$$

And we have
$$|a+(2m-1)d-a|=10.5\tag3$$

Now solve $(1)(2)(3)$ to get $a,d,2m$.

From $(1)-(2)$, we have $d=\frac 6m$. From $(3)$, we have $(2m-1)|\frac 6m|=10.5\Rightarrow m=4$. Finally, from $(1)$, we have $d=a=\frac 32$.