Question

plzz solve q.8 ............
plss​

Answer 1

Answer:

4,5,6

Step-by-step explanation:

$let \: the \: 3 \: no.s \: be \: x, \: x+1, \: and \: x+2 \\ \\ as \: per \: the \: question \\ \\ = > x(x + 1) + x(x + 2) + (x + 1)(x + 2) + = 74 \\ \\ = > {x}^{2} + x + {x}^{2} + 2x + {x}^{2} + 3x + 2 = 74 \\ \\ = > 3{x}^{2} + 6x = 72 \\ \\ = > {x}^{2} +2 x - 24 = 0 \\ \\ = > (x + 6)(x - 4) = 0 \\ \\ =>x = - 6 \: or \: x = 4 \\\\ therefore \: we \: take \: x = 4 \: \\ since \: x = - 6 \: is \: not \: a \: natural \: number\\ \\ the \: numbers \: are \: 4, \: 5 \: and\: 6, \: \:$