# The sum of 6 consecutive odd numbers is 20. What is the fourth number in this sequence?

Jan 2, 2017

There is no such sequence of $6$ consecutive odd numbers.

#### Explanation:

Denote the fourth number by $n$.

Then the six numbers are:

$n - 6 , n - 4 , n - 2 , \textcolor{b l u e}{n} , n + 2 , n + 4$

and we have:

$20 = \left(n - 6\right) + \left(n - 4\right) + \left(n - 2\right) + n + \left(n + 2\right) + \left(n + 4\right)$

$\textcolor{w h i t e}{20} = \left(n - 6\right) + 5 n$

$\textcolor{w h i t e}{20} = 6 n - 6$

Add $6$ to both ends to get:

$26 = 6 n$

Divide both sides by $6$ and transpose to find:

$n = \frac{26}{6} = \frac{13}{3}$

Hmmm. That is not an integer, let alone an odd integer.

So there is no suitable sequence of $6$ consecutive odd integers.

$\textcolor{w h i t e}{}$
What are the possible sums of a sequence of $6$ consecutive odd numbers?

Let the average of the numbers be the even number $2 k$ where $k$ is an integer.

Then the six consectuvie odd numbers are:

$2 k - 5 , 2 k - 3 , 2 k - 1 , 2 k + 1 , 2 k + 3 , 2 k + 5$

Their sum is:

$\left(2 k - 5\right) + \left(2 k - 3\right) + \left(2 k - 1\right) + \left(2 k + 1\right) + \left(2 k + 3\right) + \left(2 k + 5\right) = 12 k$

So any multiple of $12$ is a possible sum.

Perhaps the sum in the question should have been $120$ rather than $20$. Then the fourth number would be $21$.