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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)

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  • $\begingroup$ Check the numbers - this one doesn't seem to give sensible number of terms... $\endgroup$ – Macavity Mar 17 '16 at 11:30
  • $\begingroup$ @Macavity please re-read I have edited, there was a mistake. $\endgroup$ – Slavi Mar 17 '16 at 11:32
  • $\begingroup$ @AhmedAlaa: Let $a_i$ be the $i$th term. Does "its terms is even" mean that the number of the terms is even? If so, let it be $2m$. Then, does "total of it's even terms $=30$" mean that $a_2+a_4+a_6+\cdots +a_{2m}=30$ ? Also, does "total of it's odd terms $=24$" mean that $a_1+a_3+a_5+\cdots +a_{2m-1}=24$ ? (by the way, if all of these are correct, the question can be solved.) $\endgroup$ – mathlove Mar 19 '16 at 7:20
  • $\begingroup$ Yeah "its terms is even" means that the number of the terms is even, I can't see good in this, please answer the question. $\endgroup$ – Slavi Mar 19 '16 at 10:59
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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 $d=a=\frac 32$.

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