SOLUTION: The sum of five consecutive integers is 40. What is the smallest of the integers?
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Question 260234: The sum of five consecutive integers is 40. What is the smallest of the integers?
Found 2 solutions by richwmiller, ptfile:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
n+n+1+n+2+n+3+n+4=40
5n+10=40
5n=30
n=6
the smallest is 6
6,7,8,9,10
Answer by ptfile(81) (Show Source): You can put this solution on YOUR website!
Let n = 5
d = 1
40 = the sum of 5 consecutive integers
Sn = n/2(2a1+(n-1)d)
40 = 5/2(2a1+(5-1)1)
40 = 5/2(2a1+4)
40 = (5/2) 2(a1+2)
40 = 5(a1+2)
8 = a1+2
6 = a1
The smallest number is 6.
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