Question 122965
The general expression for consecutive multiples of 5 is 5N, 5(N+1), 5(N+2), and so on, where N is some unspecified integer. Find four consecutive multiples of 5 ssuch that 6 times the first is 40 greater than 2 times the sum of the second and fourth. 
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EQUATION:
6*5n = 2[(5(n+1)+5(n+3)]+40
30n = 2[10n+20]+40
30n = 20n+80
10n = -80
n = -8
5n=-40
5(n+1)=-35
5(n+2)=-30
5(n+3)=-25
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Comment: I changed your problem from "40 times greater than" to 
40 greater than.
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Cheers,
Stan H.