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 

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