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)

  • $\begingroup$ 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

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 $d=a=\frac 32$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.