Sum of all $n$ arithmetic means is equal to $n$ times single arithmetic mean.

Question

My teacher asked us to note a property which states that if there are two APs (Arithmetic Progression) such as $a, x_1, x_2, x_3, \dots, x_n, b$ and $a, y, b$, then $ny = x_1 + x_2 + x_3 + x_4+\dots+ x_n.$

I cannot visualize this, I mean how??? I understand that $x_1 + x_2 + x_3 + \cdots + x_n=ny$ where $y$ is some number, but I don't understand why would that number fit so gracefully between $a$ and $b$ such that $b - y = y - a?$

Thanks.

Answer 1

Let $x_i=a+id\;$ and $\;b=a+(n+1)d$, where $d$ is the common difference of the first A.P. Observe, $$\sum_{i=1}^n\;x_i=\sum_{i=1}^n\; a+id=na+\frac{n(n+1)}{2}d=n\left(\frac{2a+(n+1)d}{2}\right)=n\left(\frac{a+b}{2}\right)=ny .$$

Answer 2

Use the fact the terms equidistant from the ends sum up to twice the arithmetic mean, AM.

Thus, the sum of all the arithmetic means is the product of the number of arithmetic means and the AM of the end terms.

Answer 3

$\mathbf{n}$ ARITHMETIC MEANS BETWEEN TWO GIVEN NUMBERS

Let $a$ and $b$ be two given numbers. Let $x_{1}, x_{2}, \ldots, x_{n}$ be the $n$ arithmetic means between $a$ and $b$. Then $a, x_{1}, x_{2}, \ldots, x_{n}, b$ are in A.P. In this A.P., the first term is $a$, the last term is $b$ and the number of terms is $(n+2)$. Let $d$ be the common difference of the A.P. Then, $$ \begin{array}{l} b=a+(n+2-1) d \Rightarrow d=\left\{\frac{(b-a)}{(n+1)}\right\} \\ x_{1}=\left[a+\frac{(b-a)}{(n+1)}\right] ; x_{2}=\left[a+\frac{2(b-a)}{(n+1)}\right] \\ x_{n}=\left[a+\frac{n(b-a)}{(n+1)}\right] \quad n[48)= \end{array} $$ are the required arithmetic means between $a$ and $b$.

For the sum of $n$ arithmetic means, see this post Number of Arithmetic Means inserted between 2 quantities