The average of five consecutive natural numbers is m. If the next three natural numbers are also included, how much more than m will the average of these 8 numbers be ?

Guest Feb 5, 2022

#1**+1 **

Let the five consecutive natural numbers be a, b, x, d, e where a = x - 2, b = x - 1, x = x, d = x + 1 and e = x + 2.

That's like (x - 2), (x - 1), x, (x + 1), (x + 2). The average of those natural numbers is \(\frac{(x - 2) + (x - 1) + x + (x + 1) + (x + 2)}{5}\) (to find the average you must add the n natural numbers together and divide by the total number of the consecutive numbers m).

\(\frac{(x - 2) + (x - 1) + x + (x + 1) + (x + 2)}{5} = \frac{5x}{5} = m\), this brings us to m = x. The next three natural numbers are f, g, h where f = x + 3, g = x + 4, h = x + 5, or better to say x + 3, x + 4, x + 5.

\(\frac{(x - 2) + (x - 1) + x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5)}{8}\) -

Now shorten the - 2 and 2, -1 and 1 \(\Rightarrow \frac{x + x + x + x + x + (x + 3) + (x + 4) + (x + 5)}{8} = \frac{8x + (3 + 4 + 5)}{8} = \frac{8x + 12}{8} = x + \frac{12}{8}\).

We know that m = x, so \(x + \frac{12}{8} = m + 1.5\).

Therefore, the average of the 8 consecutive natural numbers is 1.5 more than m.

Straight Feb 5, 2022