Let three consecutiveintegers be x, x + 1, x+ 2. According to the question,
2x + 3(x+ 1) + 4(x+ 2) = 74
2x+ 3x + 3 + 4x + 8 = 74
9x+ 11 = 74
On transposing 11 to R.H.S, we obtain
9x= 74 - 11
9x= 63
On dividing both sides by 9, we obtain
\(\frac{9x}{9} = \frac{63}{9}\)
x= 7
x+ 1 = 7 + 1 = 8
x+ 2 = 7 + 2 = 9
Hence, the numbers are 7, 8, and 9.
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