Question

three consecutive odd natural number are such that the product of the first and third is greater than four times the middle by 1. find the number.​

Answer 1

Answer:

$\boxed{3,\:5,\:7}$

Step-by-step explanation:

$Let\:1st\:odd\:number\:be\:x,second\:x+2\:and\:third\:x+4$

$x(x+4)=4(x+2)+1$

$x^{2} +4x=4x+8+1$

$x^{2} =9$

$x=\sqrt{9}$

$x=3\:and\:x=-3(neglect\:this)$

$The\:Numbers\:are\:3$

$5$

$and\:7$

Answer 2

$<font color="purple"><b>$

$\huge{\boxed{\boxed{\mathbf{Answer:-}}}}$

let,

the no. be ,

$x \: \: (x + 2) \: \: (x + 4)$

then,

$= > x(x + 4) = 4(x + 2) + 1$

$= > {x}^{2} + 4x = 4x + 8 + 1$

$= > {x}^{2} = 9$

$= > x = 3$

the three odd numbers are =

5, 7, 9

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