Question

when the smallest of three consecutive odd integers is added to four times the largest it produces a result 729 more than four time the middle integer find the numbers and check your answer​

Answer 1

Step-by-step explanation:

Let the three integers be x, x+2 and x+4

Now from the given information, the equation we get is

x+4(x+4)=4(x+2)+729

⇒x+4x+16=4x+8+729

⇒5x−4x=737−16

⇒x=721

So the three consecutive odd integers are,

x=721

x+2=723,

and x+4=725

Therefore, the numbers are 721,723,725.

Answer 2

S O L U T I O N :

Let, the first consecutive even integer be, x.

Then, the second consecutive even integer be, x + 2.

So, the third consecutive even integer be, x + 4.

It is Given that,

The smallest of three consecutive odd integers is added to four times the largest it produces a result 729 more than four time the middle integer.

As per question, we get,

➮ x + 4 ( x + 4 ) = 4 ( x + 2 ) + 729

➮ x + 4x + 16 = 4x + 8 + 729

➮ x + 4x - 4x = 729 + 8 - 16

➮ 5x - 4x = 737 - 16

➮ x = 721

Therefore, We got the value of, x = 721.

Hence,

The first consecutive even integer , x = 721

Then, the second consecutive even integer , x + 2 = 721 + 2 = 723

So, the third consecutive even integer be, x + 4 = 721 + 4 = 725