Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am working on the following question:

Let us say we have $n$ consecutive numbers labelled $x_1$ to $x_n$. I want to find the average of these. The quickest way to do this seems to be $$AVERAGE=\frac{x_1 + x_n}{2}\tag1$$

The other way to do this is $$AVERAGE=\frac{x_1+x_2+x_3...+x_n}{n}\tag2$$

I stumbled upon equation (1) without any proof of it, but simply started using it as it seems to work, and later I proved equation (1) by converting the problem into one of coordinate geometry. I want to know if this shortcut (i.e. equation 1) is legitimate, since I am not absolutely sure about my proof.

share|cite|improve this question

Yes, it’s legitimate. Let $y=x_1+x_n$. Notice that $x_2+x_{n-1}=y$ as well, since $x_2=x_1+1$ and $x_{n-1}=x_n-1$. In fact, each column on the righthand side of the equals signs in the array below sums to $y$:

$$\begin{array}{cc} S&=&x_1&+&x_2&+&x_3&+&\ldots&+&x_{n-2}&+&x_{n-1}&+&x_n\\ S&=&x_n&+&x_{n-1}&+&x_{n-2}&+&\ldots&+&x_3&+&x_2&+&x_1\\ \hline 2S&=&y&+&y&+&y&+&\ldots&+&y&+&y&+&y \end{array}$$

Thus, $2S=ny$, $S=\dfrac{ny}2$, and the average $\dfrac{S}n=\dfrac{ny}{2n}=\dfrac{y}2$, which is exactly your shortcut formula.

In fact the shortcut will work whenever the numbers $x_1,\dots,x_n$ are in arithmetic progression.

share|cite|improve this answer

Let's say the consecutive numbers start with $a$, so that $$x_1=a,\quad x_2=a+1,\quad\ldots,\quad x_n=a+(n-1)$$ Then $$\begin{align*} x_1+x_2+\cdots+x_n&=a+(a+1)+\cdots+(a+(n-1))\\ &=(\underbrace{a+a+\cdots+a}_{n\text{ times}})+\bigg[1+2+\cdots+(n-1)\bigg]\\\\\\\\ &=na+\frac{n(n-1)}{2} \end{align*}$$ (where we have used the standard formula for the sum of the first $k$ integers) so that $$\frac{x_1+x_2+\cdots+x_n}{n}=a+\frac{n-1}{2}=\frac{2a+(n-1)}{2}=\frac{a+(a+(n-1))}{2}=\frac{x_1+x_n}{2}$$

share|cite|improve this answer

In general, if you have an arithmetic progression, the average of $n$ consecutive terms is the average of the first and the last one. The proof is as below.

Consider an arithmetic progression with the first term being $a$ and the common difference being $d$. The $n^{th}$ term is given by $a+(n-1)d$, i.e., the first $n$ terms are $$a,a+d,a+2d,\ldots, a+(n-1)d.$$ Then the average of the first $n$ terms is \begin{align} \dfrac{a+(a+d) + (a+2d) + \cdots + (a + (n-1)d)}n & = \dfrac{na + \dfrac{n(n-1)}2d}n\\ & = \dfrac{a+(a+(n-1)d)}2\\ & = \dfrac{\text{First} + \text{Last}}2 \end{align}

share|cite|improve this answer

Alicia and her friend Bob both got jobs at the same place.

On the first day, Alicia earned $a$ dollars. She got a raise of $d$ dollars the next day so earned $a+d$ dollars. And she got another raise of $d$ dollars the day after that, and so on. She worked a total of $n$ days, getting a raise of $d$ dollars each day. Let us suppose that on the $n$-th day, which was the last, she earned $b$ dollars.

Bob started at the same time as Alicia, and earned $b$ dollars the first day. But he was pretty incompetent, and his wage the next day was decreased by $d$ dollars, and the next day again by $d$ dollars, and so on. On his $n$-th and last day, he earned $a$ dollars.

It so happens that Alicia and Bob live together. Note that between them, they earned $a+b$ dollars the first day. The same is true for the next day, and the day after that, and so on. For each day Alicia's income rises by $d$, and Bob's falls by $d$, so their combined daily income stays constant.

Between them, they earned $n(a+b)$ dollars. Alicia earned half of this, so she earned $\dfrac{n(a+b)}{2}$ dollars. So her average wage was $\dfrac{a+b}{2}$ per day.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.