Hi
Let x and (x+2) and (x+4)
represent the first, second and third of three consecutive odd numbers
Question States***
(1/7) x + 1/3(x+2) +1/5(x+4) = 63
solving for x
Multiplying all terms by 105, so as all denominators = 1
15 + 35(x+2) + 21(x+4) = 63*105
71x + 154 = 6615
71x = 6461
x = 91 Three consecutive even numbers are 91,93,95
CHECKING our Answer***
1/7*91 + 1/3*93 + 1/5*95 = 13 + 31 + 19 = 63