A conjecture regarding Arithmetic Progressions I have a conjecture in my mind regarding Arithmetic Progressions, but I can't seem to prove it. I am quite sure that the conjecture is true though.
The conjecture is this: suppose you have an AP (arithmetic progression):
$$a[n] = a[1] + (n-1)d$$
Now, suppose our AP satisfies the property that the sum of the first $n$ terms of our AP is equal to the sum of the first $m$ terms:
$$S[n] = S[m]$$
but $n \neq m$. I want to prove two theorems:

*

*The underlying AP $a[n]$ must be symmetric with respect to the point at which it becomes zero.

*$S[n + m] = 0$
A Numerical Example
Consider the AP:
$$a[n] = 4 - n = (3, 2, 1, 0, -1, -2, -3)$$
This is an AP with common difference $d = -1$ and first term $a[1] = 3$:
Here is the MATLAB plot of this AP. As you can see in the plot, our AP is symmetric with respect to the point $n = 4$:
$$a[4-1] = -a[4+1] = 1$$
$$a[4-2] = -a[4+2] = 2$$
$$a[4-3] = -a[4+3] = 3$$
Now, here is the sum of our AP:
$$S[n] = (3,5,6,6,5,3,0)$$
Here is the MATLAB plot of the summation. You can clearly see that:
$$S[1] = S[6] = 3$$
$$S[2] = S[5] = 5$$
$$S[3] = S[4] = 6$$
and you can also see that:
$$S[1 + 6] = S[7] = 0$$
$$S[2 + 5] = S[7] = 0$$
$$S[3 + 4] = S[7] = 0$$
Can you please help me out with this problem? Any guidance will be very welcome. I am actually an Engineering student, so my Pure Math skills are not that strong.
Thank you!
 A: Ok. Let the first term be $a$ and the ratio $r$. Then the sum of the first $n$ terms is $an+\frac{(n-1)n}{2}$ and the sum of the first $m$ terms is $am+\frac{r(m-1)m}{2}$.
Suppose those are equal. Then $$an+\frac{r(n-1)n}{2}=am+\frac{r(m-1)m}{2}$$ which is equivalent to $$a(n-m)=r\frac{m^2-n^2-(m-n)}{2}=r\frac{(m-n)(m+n-1)}{2}=r\frac{(n-m)(1-m-n)}{2}$$
Because $n\neq m$, we have $a=\frac{r(1-m-n)}{2}$.
Then, the sum of the first $m+n$ terms is $$a(m+n)+\frac{r(m+n-1)(m+n)}{2}=(m+n)\times\frac{r(1-m-n)}{2}+\frac{r(m+n-1)(m+n)}{2}=-\frac{r(m+n-1)(m+n)}{2}+\frac{r(m+n-1)(m+n)}{2}=0$$
To prove it is symetric, lets find when it becoems $0$.
If the $k^{th}$ term is 0, then $a+kr=0$, so $\frac{r(1-m-n)}{2}+kr=0$, leading to $k=\frac{m+n-1}{2}$. Notice that for this to happen, $m+n$ must be odd. Anyhow, from this point, to prove the symmetry, just prove that the $|(\frac{m+n-1}{2}-x)^{th}$ term$|$ is equal to $|(\frac{m+n-1}{2}+x)^{th}$ term $|$ term, using the formula (when $m+n$ is odd)
A: If $S_n=S_m$ with $n\ne m$, i.e. $na_1+\frac{n(n-1)}{2}d=ma_1+\frac{m(m-1)}{2}d$, this means that $a_1(n-m)+\frac{(n-m)(m+n-1)}{2}d=0$ . As $n\ne m$, we can cancel $n-m$ and conclude:
$$a_1+\frac{m+n-1}{2}d=0$$
Now:
$$\begin{array}{rcl}S_{m+n}&=&a_1(m+n)+\frac{(m+n)(m+n-1)}{2}d\\&=&(m+n)\left(a_1+\frac{m+n-1}{2}d\right)\\&=&(m+n)\cdot 0\\&=&0\end{array}$$
As for the symmetry, let's define $k=\frac{m+n+1}{2}$ and the above formula means that $a_1+(k-1)d=0$. Now, let $i$ be a number such that $k+i$ and $k-i$ are both integers:
$$a_{k+i}=a_1+(k+i-1)d=a_1+(k-1)d+id=id$$
$$a_{k-i}=a_1+(k-i-1)d=a_1+(k-1)d-id=-id$$
so it follows that $a_{k-i}=-a_{k+i}$ - thereby proving the symmetry.
What $i$ will be in the above proof depends on whether $m+n$ is odd - in which case $k$ is an integer and $i$ is an integer, or $m+n$ is even, in which case $k$ is a "half-integer" (i.e. an integer $+\frac{1}{2}$), and $i$ is also a "half-integer".
The sequence you have given as an example is the example of the former case:
$$3,2,1,0,-1,-2,-3,\ldots$$
and e.g. $S_1=S_6=3$. This means that, $k=\frac{1+6+1}{2}=4$ is an integer, and $a_k=a_1+(k-1)d=0$. (Indeed, $a_4=0$.) Moreover, taking $i=1,2,3,\ldots$, we conclude $a_5=-a_3,a_6=-a_2,a_7=-a_1$.
The example of the latter case is the sequence:
$$5,3,1,-1,-3,-5,-7\ldots$$
where e.g. $S_1=S_5=5$ and $k=\frac{1+5+1}{2}=3\frac{1}{2}$ - a "half-integer". By taking $i=\frac{1}{2},\frac{3}{2},\frac{5}{2},\ldots$ we get that $a_4=-a_3, a_5=-a_2, a_6=-a_1$ - even though this sequence never takes the value $0$.
