The sum of eight consecutive (one after the other) numbers is 88. What are the numbers?

Question

Levi Adams · Accepted Answer

If we call the first number, an unknown, #n#, then the next consecutive numbers are #n+1,n+2. . . n+7#. Thus, when we add all the terms, from #n# to #n+7# together, and set that equal to #88#, we get: #8n+28=88# Note that #1+2+3+4+5+6+7=28#. This gives us #8n=60# #n=15/2# Note that this is not an integer, which leads us into shaky territory: its hard to define #15/2,17/2,19/2. . .# as "consecutive numbers," per se. A more apt description of this would be that the sum of these #8# numbers, all of which are #1# away from the next highest number, have a sum of #88#. #15/2+17/2+19/2+21/2+23/2+25/2+27/2+29/2=88#

Gabriel Agustin · Answer

undefined Let's represent the smallest of the eight consecutive numbers as $ x $. Then the next seven consecutive numbers would be $ x + 1 $, $ x + 2 $, $ x + 3 $, $ x + 4 $, $ x + 5 $, $ x + 6 $, and $ x + 7 $.

The sum of these eight consecutive numbers is given as 88:

$$ x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6) + (x + 7) = 88 $$

Solve this equation for $ x $ to find the smallest number in the sequence. Then, you can determine the other consecutive numbers by adding the corresponding values of $ x $.