Find $k$ in arithmetic progression knowing $a_4$, $n=10$ and knowing fact of $S_{k\,\text{last}}$

Question

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.

Answer 1

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. $a_i=i-4$.

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METHOD 1

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

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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 $11k-T$.

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

Answer 2

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 $k=1$)

Answer 3

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 $1\le k\le 10$, $k=8$.

Answer 4

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.