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Let the three consecutive numbers be x + 1,x + 2,x + 3.
Given that sum of second and the third number exceeds the first term by 14.
= > x + 2 + x + 3 = x + 1 + 14
= > 2x + 5 = x + 15
= > 2x - x = 15 - 5
= > x = 10.
Then,
= > x + 1 = 11
= > x + 2 = 12
= > x + 3 = 13.
Therefore, the three numbers are 11,12,13.
Hope this helps!