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Let's consider arithmetic progression with integer numbers.

Arithmetic progression sum $S_n = (a_1 + a_n)*n/2$, where $a_n=a_1+d(n-1) $

So $ S_n = (2*a_1 + d(n-1))*n/2 = a_1*n + d(n-1)*n/2$

I cannot understand, why it always happens that $d(n-1)*n/2$ is always an integer number? So that $S_n$ is also always an integer.

Besides, everything seems clear with even number of progression members: 1 + 2 + 3 + 4 + 5 + 6 = (1+6) + (2+5) + (3+4) = (1+6)* (6 members / 2)

But if number of progression members is not even (1 + 2 + 3), it is unclear why $S_n = (a_1 + a_n)*n/2$ formula works perfectly! Because (3 members / 2) is not an integer!

Code Complete
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    With an odd number of terms, there are an even number of steps between the first and last term, so they have the same parity (both odd or both even). Therefore $a_1+a_n$ will always be even when $n$ is odd. – Jaap Scherphuis Jun 13 '18 at 15:47
  • @ Jaap Scherphuis "so they have the same parity (both odd or both even)" - Sorry, could you please give a hint, "they" means what? Your answer is very interesting, and I'm trying to understand it fully. – Code Complete Jun 13 '18 at 15:52
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    If there are an odd number of terms, then the first and last term, i.e $a_1$ and $a_n$, are both odd or both even. This is because there are an even number of steps between them (namely $n-1$ steps) making their difference even too (that difference being $d(n-1)$ ). Therefore their sum, $a_1+a_n$, is even when $n$ is odd. – Jaap Scherphuis Jun 13 '18 at 19:40

4 Answers4

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$\frac{d(n-1)n}{2}$ is always an integer because $\frac{(n-1)n}{2}$ is always an integer. This is simply because $n$ and $n-1$ differ by $1$, and so atleast one of them must be even.

Manan
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$S_n = \dfrac{n(first + last)}{2}$

where $first = a_1$ and $last = a_n = a_1+(n-1)d$

Note that, if $n$ is odd, then

$$first + last = 2a_1+(n-1)d$$

which is even because $2a_1$ is even and $(n-1)d$ is even.

Steven Alexis Gregory
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If $n=2m$, then: $$S_{2m}=\require{cancel}\frac{a_1+a_{2m}}{\cancel{2}}\cdot \cancel{2}m=(a_1+a_{2m})\cdot m.$$ If $n=2m-1$, then: $$S_{2m-1}=\require{cancel}\frac{a_1+a_{2m-1}}{2}\cdot (2m-1)=\frac{\cancel{2}(a_1+d(m-1))}{\cancel{2}}\cdot (2m-1)=a_m\cdot (2m-1).$$

farruhota
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This assumes we already know it that the formula is for an AP*.

We should only be concerned with the case where both $(a_1+a_n)$ and $n$ are odd, otherwise clearly $S_n$ will be an integer.

For $(a_1+a_n)$ to be odd, either $a_1$ is odd and $a_n$ is even, or the other way round. In both cases, this requires the number of terme $n$ to be even. This is shown below.

Let $d$ be the common difference of the AP. Consider the following:

  • If $a_1$ is odd and $d$ is odd, the parity of the AP is $OEOEO\cdots$, i.e. $a_n$ is even only for even $n$.
  • If $a_1$ is odd and $d$ is even, the parity of the AP is $OOOOO\cdots$, i.e. $a_n$ is odd, so not a possibility, as both $a_1$ and $a_n$ are odd.
  • If $a_1$ is even and $d$ is odd, the parity of the AP is $EOEOE\cdots$, i.e. $a_n$ is odd only for even $n$.
  • If $a_1$ is even and $d$ is even, the parity of the AP is $EEEEE\cdots $, i.e. not a possibility.

Hence it is not possible for both $(a_1+a_n)$ and $n$ to be odd at the same time, so at least one of them must be even.

As such, their product is even and $S_n$ is an integer. $\blacksquare$.

*If we did not know that $$S_n=(a_1+a_n)\cdot \frac n2$$ refers to an AP in the first, then we would have to prove that first.

Here's one approach of doing that.

$$\begin{align} &2S_{n-1}&&=(n-1)\left(a_1+a_{n-1}\right)\\ &2S_{n}&&=\qquad n\left(a_1+\ a_{n}\; \right)\\ &2S_{n+1}&&=(n+1)\left(a_1+a_{n+1}\right)\\\\ &2(S_{n+1}-S_n)=2a_{n+1}&&=a_1+(n+1)a_{n+1}-na_n\\ & \qquad\qquad\qquad 0 &&=a_1+(n-1)a_{n+1}-na_n\tag {1}\\\\ &2(S_{n}-S_{n-1})=2a_{n}&&=a_1+na_{n}-(n-1)a_{n-1}\\ & \qquad\qquad \qquad 0 &&=a_1+(n-2)a_{n+1}-(n-1)a_n\tag {2}\\\\ (1)-(2):\qquad &\qquad\qquad\qquad 0&&=0+(n-1)a_{n+1}-(2n-2)a_n+(n-1)a_{n-1}\\ &\qquad\qquad\qquad 0&&=0+a_{n+1}-2a_n+a_{n-1}\\ &\qquad\qquad\qquad 2a_n&&=a_{n+1}+a_{n-1}\\ \end{align}$$ Hence the series is an AP. $\blacksquare$

Hypergeometricx
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