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

1 Answer
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, color(blue)(n), n+2, n+4#

and we have:

#20 = (n-6)+(n-4)+(n-2)+n+(n+2)+(n+4)#

#color(white)(20) = (n-6)+5n#

#color(white)(20) = 6n-6#

Add #6# to both ends to get:

#26 = 6n#

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

#n = 26/6 = 13/3#

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

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

#color(white)()#
What are the possible sums of a sequence of #6# consecutive odd numbers?

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

Then the six consectuvie odd numbers are:

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

Their sum is:

#(2k-5)+(2k-3)+(2k-1)+(2k+1)+(2k+3)+(2k+5) = 12k#

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#.