Question 333638
he general expression for consecutive multiples of seven is 7N, 7(N+1), 7(N+2), etc. 
Find three consecutive multiples of seven such that 3 times the first exceeds
 twice the third by 14.
:
3(7N) = 2(7(n+2) + 14
21N = 14(N+2) + 14
21N = 14N + 28 + 14
21N - 14n + 42
7N = 42, is the 1st multiple
then
42, 49, 56  are the 3 multiples