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

N, n+1, n+2  are the numbers.

n/4+(n+1)/5=(n+2)/7-5 make a common denominator of 140

(35n+28n+28)/140=(20n+40-700)/140  multiply both sides by 140

35n+28n+28=20n+40-700

63n+28=20n-660

43n+28=-660

43n=-688

n=-16

since we started with n, n+1, and n+2 the numbers are:

-16, -15, -14