find three consecutive natural numbers such that the sum of the first and the second is 39 more than the third number
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0
N1 = 40
N2 = 41
N3 = 42
N1 + N2 = 81 - N3 = 39
N2 = 41
N3 = 42
N1 + N2 = 81 - N3 = 39
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Let the three numbers be: x-1, x and x+1.
-Therefore, x-1 + x + x+1 =3x = 39
-Hence x, the middle number is 13.
-The other two numbers are 12 and 14.:
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