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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.

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4 Answers 4

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.

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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 $$\frac{x_1+x_2+\cdots+x_n}{n}=a+\frac{n-1}{2}=\frac{2a+(n-1)}{2}=\frac{a+(a+(n-1))}{2}=\frac{x_1+x_n}{2}$$

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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}

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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.

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