Find $k$ in arithmetic progression knowing $a_4$, $n=10$ and knowing fact of $S_{k\,\text{last}}$ 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.
 A: 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 $k=1$)
A: 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 $11k-T$. 
$$\begin{align}
5(\overbrace{T-4k}^{S_{k\text { first}}})&=\overbrace{(11k-T)-4k}^{S_{k\text{ last}}}\\
6T&=27k\\
2\cdot \frac{k(k+1)}2&=9k\\
\because k\neq 0\therefore \qquad\qquad k&=8\qquad\blacksquare
\end{align}$$
A: 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 $1\le k\le 10$, $k=8$.
A: 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.
