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# Algebra

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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 ?

Feb 5, 2022

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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.

Feb 5, 2022
edited by Straight  Feb 5, 2022