SOLUTION: Find four consecutive multiples of five such that twice the sum of the two biggest exceeds twice the smallest by ten.

Question 250740: Find four consecutive multiples of five such that twice the sum of the two biggest exceeds twice the smallest by ten.
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!
Find four consecutive multiples of five such that twice the sum of the two biggest exceeds twice the smallest by ten.


5k     = smallest multiple of 5
5(k+1) = next to smallest multiple of 5
5(k+2) = next to biggest multiple of 5
5(k+3) = biggest multiple of 5

>>twice the sum of the two biggest... 

sum of the two biggest = 5(k+2) + 5(k+3)

twice the sum of the two biggest = 2[5(k+2) + 5(k+3)]

...exceeds twice the smallest by ten.<<

twice the smallest = 2(5k)

2[5(k+2) + 5(k+3)] = 2(5k) + 10

Can you solve that for k?  If not post again asking how.

Solution: k = -4

smallest multiple of 5         = 5k     = 5(-4)   =         -20
next to smallest multiple of 5 = 5(k+1) = 5(-4+1) = 5(-3) = -15
next to biggest multiple of 5  = 5(k+2) = 5(-4+2) = 5(-2) = -10
biggest multiple of 5          = 5(k+3) = 5(-4+3) = 5(-1) =  -5

So they are -20, -15, -10, -5.    

Edwin

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