Question

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

Answer 1

.(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