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I know that an arithmetic serie has $10$ terms and some more things:

$$a_4=0$$ $$\;\quad\qquad\qquad\qquad n= 10 \quad\text{As I said above}$$ $$S_{k\,\text{last}} = 5S_{k\,\text{first}}$$

In other words the last line says that

Sum of the $k$ last terms is $5$ times bigger than sum of the $k$ first terms.

My problem is that I don't know almost anything:

Not $a_1$ (the first term of the sequence), not $d$ (common difference, I'm supposed to not find it before).

So this it seems difficult, maybe I'm missing something. I can't unfortunately show more effort, I don't know what to do. My last effort was to traduce the question to mathematical equations as I showed at the beginning of the question with the $3$ central sentences.

A hint could help. With that I mean that maybe with one I could see the trick of this question and resolve it.

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  • $\begingroup$ Should your last equation read $5\,S_{k\;first}=S_{k\;last}$? Is that meant to hold for all $k$? so the last term is five times the first, next to last term is five times the second, and so on? Doesn't seem possible (unless all terms are $0$ of course). $\endgroup$ – lulu Apr 8 '16 at 13:58
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    $\begingroup$ You know that the terms are $-3d, -2d, -d, 0, d, 2d, 3d, 4d, 5d, 6d$. Or have you misstated the question? So the only way the last $k$ terms can be 5x bigger than the first $k$ is if $d=0$. $\endgroup$ – almagest Apr 8 '16 at 14:00
  • $\begingroup$ @lulu No. I need to find a $k$. It could be that $S_2$ = $a_9+a_{10}$, or maybe it is $S_4$ = $a_7+a_8+a_9+a_{10}$ or some few other options possible. I need to know which of these cases is the correct one. $\endgroup$ – Pichi Wuana Apr 8 '16 at 14:01
  • $\begingroup$ @almagest I'm resolving exercise a from the question. On the following exercise connected to the same question (b) they give us $d$, so I suppose they are telling me I don't need to find $d$. $\endgroup$ – Pichi Wuana Apr 8 '16 at 14:04
  • $\begingroup$ You can't determine $d$ from this data, but you can determine $k$ (assuming that $d\neq 0$). Just go by trial and error on $k=\{1,2,\cdots\}$. $\endgroup$ – lulu Apr 8 '16 at 14:07
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Without loss of generality we can assume $d=1$, as the objective here is to find $k$.

Then the series becomes $\lbrace -3, -2, -1, \;\;0, \;\;1,\;\; 2,\;\; 3, \;\; 4,\;\; 5, \;\;6\rbrace$, i.e. $a_i=i-4$.


METHOD 1

We want $$\begin{align} S_{k \text{ last}}&=5 S_{k\text{ first}}\\ \frac k2 \bigg[6+(7-k)\bigg]&=5\cdot \frac k2\bigg[-3+(k-4)\bigg]\\ 13-k&=5(k-7)&&\because k\neq 0\\ k&=8\qquad\blacksquare \qquad\end{align}$$


METHOD 2

Add $4$ to the original series, resulting in $\lbrace 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\rbrace$.

Let $T$=sum of first $k$ integers=$k(k+1)/2$.

Sum of last $k$ integers in this series is $11k-T$.

$$\begin{align} 5(\overbrace{T-4k}^{S_{k\text { first}}})&=\overbrace{(11k-T)-4k}^{S_{k\text{ last}}}\\ 6T&=27k\\ 2\cdot \frac{k(k+1)}2&=9k\\ \because k\neq 0\therefore \qquad\qquad k&=8\qquad\blacksquare \end{align}$$

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  • $\begingroup$ @Pichi Wuana - The $\blacksquare$ is known as a halmos, and indicates the solution or end of proof. FYI. $\endgroup$ – hypergeometric Apr 8 '16 at 15:22
  • $\begingroup$ Oh wow. I'm sorry for that! $\endgroup$ – Pichi Wuana Apr 8 '16 at 15:35
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    $\begingroup$ That's OK. I usually put it at the end of the solutions I post here. Will reinstate it for consistency :) Hope the solution was helpful. $\endgroup$ – hypergeometric Apr 8 '16 at 15:40
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The first two sentences say the progression is $-3d, -2d, -d, 0, d, 2d, 3d, 4d, 5d, 6d$. If $k$ were to be $1$, we would need $6d=5(-3d)$, which cannot be unless $d=0$ because they are of opposite signs. However, if you try $k=8$, you have $S_8=4d, S_{8 last}=20d$. You can't find what $d$ is, but it has to be non-zero to have a unique $k$. The key to finding it easily is to realize that $k$ can be greater than $5$ and that if it is too small the two values are of opposite signs (like for $k=1$)

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  • $\begingroup$ Do you mean that you are using just as lulu said, trial and error? $\endgroup$ – Pichi Wuana Apr 8 '16 at 14:11
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    $\begingroup$ I would call it intelligent trial and error. Once I realized $k$ had to be rather large, I know the sum of all the positive terms is $21d$. Taking off $-d$ makes the coefficient $20$, which is divisible by $5$. That makes it likely that $k=8$ is correct. There is nothing wrong with trial and error on a problem that is small enough, as this one is. $\endgroup$ – Ross Millikan Apr 8 '16 at 14:13
  • $\begingroup$ I understand. It seems a wise method to resolve it. $\endgroup$ – Pichi Wuana Apr 8 '16 at 14:17
  • $\begingroup$ I found a way without trial and error! Should I post it as an answer to my self? I never posted an answer to my own question. Or to accept the best answer? $\endgroup$ – Pichi Wuana Apr 8 '16 at 14:29
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    $\begingroup$ The faq explicitly allows answering your own question, so go ahead. You will have to wait a bit to accept it, but if it is the best answer you should do so. Often seeing some answers will help your thinking so you can find a better one. $\endgroup$ – Ross Millikan Apr 8 '16 at 14:33
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As some people said at the comments the arithmetic progression can also be rewritten as

$$-3d, -2d, -d, 0, d, 2d, 3d, 4d, 5d, 6d$$

As we can see this is also an arithmetic progression.

We know that

$$S_n = \frac n2\left(2a_1 + d(n-1)\right)$$

So I create two arithmetic progressions:

$$a_1 = -3, \quad d_a = 1 $$ $$b_1 = 6, \quad d_b = -1 $$

One for last terms and one for first terms.

We can say now that:

$$5Sa_k = Sb_k$$ $$5\frac n2(-6+k-1) = \frac n2(12-(k-1))$$

We cancel out $\frac n2$ on both sides. $$5(-6+k-1) = (12-(k-1))$$ $$5(-7+k) = 12-k+1$$ $$-35 + 5k = 12 - k +1$$

$$-48 = -6k$$

From here we find $k=8$.

A trick for resolving this type of questions is to write the arithmetic progression even if you don't think you have the necessary information, as in the top.

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We can set $a_j=(j-4)d$ for $j\in\{1,2,\cdots,10\}$. Then $$ S_{k\text{ first}}= a_1 + a_2+ \cdots + a_k =\sum_{i=1}^k (i-4)d =\left(\frac{k(k+1)}{2}-4k\right)d $$ and $$ S_{k\text{ last}}= \sum_{i=11-k}^{10}(i-4)d = \sum_{i=1}^k (7-i)d =\left(\frac{-k(k+1)}{2} +7k\right)d. $$ Since $d\ne 0$, $$ \left(\frac{-k(k+1)}{2} +7k\right)=5\left(\frac{k(k+1)}{2}-4k\right) $$ and its roots are $k=0$ or $k=8$. Since $1\le k\le 10$, $k=8$.

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  • $\begingroup$ Why is $a_n = (n-4)d$? $\endgroup$ – Pichi Wuana Apr 8 '16 at 14:22
  • $\begingroup$ @PichiWuana Because you defined $d$ as the common difference of $a_n$, and $a_4=0$ was given. $\endgroup$ – choco_addicted Apr 8 '16 at 14:24
  • $\begingroup$ You might want to use something other than $n$ for the index as it has been specified that $n=10$ (total number of terms). $\endgroup$ – hypergeometric Apr 8 '16 at 14:51

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