Why arithmetic progression formula $S_n = (a_1 + a_n)*n/2$ works with uneven number of integer members?

Question

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!

Answer 1

$\frac{d(n-1)n}{2}$ 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.

Answer 2

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

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.

Answer 3

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: $$S_{2m-1}=\require{cancel}\frac{a_1+a_{2m-1}}{2}\cdot (2m-1)=\frac{\cancel{2}(a_1+d(m-1))}{\cancel{2}}\cdot (2m-1)=a_m\cdot (2m-1).$$

Answer 4

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. $\blacksquare$.

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*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. $\blacksquare$