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Three consecutive positive even numbers are such that thrice the first number exceeds double the third by 2; then the third number is:

  1. $10$ 
  2. $14$ 
  3. $16$ 
  4. $12$
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Let the three numbers be (a - 2), a and (a + 2)

From the given condition,

3(a - 2) = 2(a + 2) + 2

On solving, we get a = 12

So the 3 numbers are 10, 12 and 14.

14 is the third number.
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