find three consecutive odd integers such that the sum of two times the first, one times the second, and three times the third is 296. list the numbers in order from smallest to largest.

We all recognize that odd numbers would be those that cannot be totally divided by two.
We can see that when we organise them in increasing order, the difference between the two is always 2.
The numbers 3, 5, 7, 9, and 11 are examples of consecutive odd numbers so because difference at all between the predecessor-successor pair is 2, as in 5 - 3 = 2 and 7 - 5 = 2.

Now, as per the question;

Let the first odd number be 'n'.

Then, the second odd number will be '(n + 2)'.

And, the third odd number will be '(n + 4)'

Now, we have to take the two times the first = 2n.

one times the second = (n + 2)

three times the third = 3(n + 4).

Sum of all three is 296.

Thus,

2n + (n + 2) + 3(n + 4) = 296

Simplifying the equation;

2n + n + 2 + 3n + 12 = 296

6n = 296 - 14

6n = 282

n = 47 (1st odd number)

n +2 = 47 + 2 = 49 (second odd number)

n + 4 = 47 + 4 = 51 (third odd number)

Thus, the three consecutive odd numbers has be calculated.

To know more about the odd numbers, here

https://brainly.com/question/2263958

#SPJ4