The sum of the 1st g and h terms of an arithmetic series are equal and g does not equal h, show that the sum of the 1st h+g terms is 0. The question I'm trying to answer is this

If the sum of the $1$st $g$ terms of an arithmetic series are the same as the sum of the $1$st $h$ terms and $h$ does not equal $g$, show that the sum of the $1$st $h+g$ terms is $0$.

First I wrote the expressions for $S_p$ and $S_q$: $S_p = \frac{p}{2}(2a + (p-1)d)$ (where $a$ is the first term and $d$ is the common difference), $S_q= \frac{q}{2}(2a + (q-1)d)$.
Then I equated them to see if I could somehow equate them and obtain values for $p$ and $q$ but I just arrived an unsolvable equation: $$2a(p-q) + d(p^2 - p - q^2 + q)$$.  
Then
 I tried directly finding $S_{p+q}$ and this is what I obtained: 
$$S_{p+q} = S_p + S_q + qpd = 2S_p + pqd$$ I've tried to take this further but I just can't obtain $S_{p+q}=0$.
I'm wondering if I can somehow simplify the above expression to obtain $0$ or if there's another way in which I have to solve the problem. 
 A: Interesting question.
In order to exploit the symmetry,  it might be easier to think of the problem as a quadratic equation in $n$ which can be graphed as a parabola, rather than using the usual AP formulas.
Note that 
$$\begin{align}
S_n&=\frac n2\big[2a+(n-1)d\big]\\
&=-\alpha n^2+\beta\\
&=-n(\alpha n-\beta)\end{align}$$
which is a quadratic in $n$ with roots $n=0, \frac {\beta}{\alpha}$, i.e. $S_0=0, S_\frac {\beta}{\alpha}=0$ and can be graphed as a downward (for $a>0, d<0$) or upward (for $a<0, d>0$) opening right parabola with $x-$axis intercepts at $x=0, \frac {\beta}{\alpha}$, and axis of symmetry $x=\frac {\beta}{2\alpha}$. Let $\frac {\beta}{2\alpha}=m$, so axis of symmetry is $x=m$. 
Given $S_h=S_g$, hence $h,g$ are equidistant from $m$.
Let $h=m-\delta$ and $g=m+\delta$ where $\delta<m$. 
Then clearly
$$h+g=2m=\frac {\beta}{\alpha}$$
Hence 
$$S_{h+g}=S_\frac {\beta}{\alpha}=0\quad\color{red}{\blacksquare}$$
NB - The only way this is possible is when $a$ and $d$ are of opposite signs. 

Addendum
Another method shown below, if you must use the standard AP formulas. This is less elegant, and does not exploit the symmetry of the problem, or make it evident.
$$\begin{align}
S_h&=S_g\\
\frac h2[2a+(h-1)]d&=\frac g2[2a+(g-1)d]\\
ah+h(h-1)\frac d2&=ag+g(g-1)\frac d2\\
a(g-h)&=\frac d2[h(h-1)-g(g-1)]\\
a&=-\frac d2\left[\frac {(h^2-g^2)-(h-g)}{h-g}\right]\\
&=-\frac d2(h+g-1)\\
S_{h+g}&=\frac {h+g}2[2a+(h+g-1)d]\\
&=(h+g)a+(h+g)(h+g-1)\frac d2\\
&=(h+g)\cdot -\frac d2(h+g-1)+(h+g)(h+g-1)\frac d2\\
&=0\quad\color{red}{\blacksquare}
\end{align}$$
A: The problem can be done without too much algebra.
Assume $g < h$ without loss of generality. First note that if $$a_1 + a_2 + \dots + a_g = a_1 + a_2 + \dots + a_h,$$ then by canceling from both sides we get $$a_{g+1} + a_{g+2} + \dots + a_h = 0.$$
At this point, it's worth noting a general rule about sums of consecutive terms in an arithmetic progression: $$a_s + a_{s+1} + a_{s+2} + \dots + a_t = (t - s + 1) a_{(s+t)/2}$$ (the middle term multiplied by the number of terms). There are any number of algebraic proofs of this, but it should be intuitively true: each term less than $a_{(s+t)/2}$ is balanced by a corresponding term that's greater than $a_{(s+t)/2}$. This identity also works in cases where $s+t$ is odd by defining $a_{i+1/2} = \frac{a_i + a_{i+1}}{2}$. 
So we have $$0 = a_{g+1} + a_{g+2} + \dots + a_h = (h-g) \cdot a_{(g+h+1)/2}$$ from which it follows that $a_{(g+h+1)/2} = 0$, and therefore
$$a_1 + a_2 + \dots + a_{g+h} = (g+h) \cdot a_{(g+h+1)/2} = 0.$$
