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Find three consecutive numbers such that the sum of one-fourth the first and one-fifth the second is five less than one-seventh the third. -10, -11 and -12 -12, -13 and -14 -14, -15 and -16
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.(n-1)/4 + n/5 = (n+1)/7 - 5 [5(n-1) + 4n]/20 = [(n+1) - 35]/7 (9n - 5)/20 = (n-34)/7 7(9n-5) = 20(n-34) 63n - 35 = 20n - 680 43n = -645 n = -15 so it's the last one
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