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.

Question

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.

Answer 1

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.

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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.

$$\begin{align} S_h&=S_g\\ \frac h2[2a+(h-1)]d&=\frac g2[2a+(g-1)d]\\ ah+h(h-1)\frac d2&=ag+g(g-1)\frac d2\\ a(g-h)&=\frac d2[h(h-1)-g(g-1)]\\ a&=-\frac d2\left[\frac {(h^2-g^2)-(h-g)}{h-g}\right]\\ &=-\frac d2(h+g-1)\\ S_{h+g}&=\frac {h+g}2[2a+(h+g-1)d]\\ &=(h+g)a+(h+g)(h+g-1)\frac d2\\ &=(h+g)\cdot -\frac d2(h+g-1)+(h+g)(h+g-1)\frac d2\\ &=0\quad\color{red}{\blacksquare} \end{align}$$

Answer 2

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 $$a_1 + a_2 + \dots + a_{g+h} = (g+h) \cdot a_{(g+h+1)/2} = 0.$$