Arithmetic Progression

**slevvio** · January 1st 2008, 03:15 PM

Hello I was having trouble with this problem and was wondering if anybody could help me with it. Show that if the sum of the first p terms of an arithmetic progression is equal to the sum of the first q terms, where p $\not=$ q, then the sum of the first p + q terms must be zero.

I do not really know where to start; I came up with this but it is not helpful:

$\frac{p}{2}(2a + (p-1)d) = \frac{q}{2}(2a + (q-1)d)$

which of course shows that p = q even though it does not.

Thanks and happy new year!

**Plato** · January 1st 2008, 04:22 PM

Look again.
The sum of the first p terms is: $\sum\limits_{k = 0}^{p - 1} {\left( {a + kd} \right)} = pa + \frac{{\left( {p - 1} \right)p}d}{2}<br />$

**Isomorphism** · January 1st 2008, 06:36 PM

We will follow what you did,

Originally Posted by slevvio

I do not really know where to start; I came up with this but it is not helpful:
$\frac{p}{2}(2a + (p-1)d) = \frac{q}{2}(2a + (q-1)d)$
which of course shows that p = q even though it does not.
Thanks and happy new year!

Not really, it says p=q could be possible, but it doesn't say p=q is the only thing possible.
What is the sum of first p+q terms??
$S = \frac{(p+q)(2a+(p+q-1)d)}2$
We want to show S=0 ,i.e $2a+(p+q-1)d = 0$ --------------(*)
Let us see why,
$\frac{p}{2}(2a + (p-1)d) = \frac{q}{2}(2a + (q-1)d)$
$2ap +dp^2 - dp = 2aq + dq^2 - dq$
$2a(p-q) +d(p^2 - q^2) -d(p-q) = 0$
$(p-q)(2a + d(p+q)-d) = 0$
Since $p \neq q$ we are forced to conclude
$2a+(p+q-1)d = 0$ which is nothing but (*).

All done

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