Determine three consecutive odd integers such that three times the middle integer is seven more than the sum of the first and third integers.

nos are n,n+2andn+4 3(n+2)=7+n+n+4 3n+6=11+2n n=5 so the nos are 5,7 and 9

To determine three consecutive odd integers, let's assign variables to represent each integer.

Let's say that the first odd integer is represented by n. Since we are looking for consecutive odd integers, the second odd integer would be n + 2, and the third odd integer would be n + 4.

According to the given condition, three times the middle integer (3 * (n + 2)) is equal to the sum of the first and third integers (n + n + 4) plus seven more.

Now we can set up the equation and solve for n:

3 * (n + 2) = (n + n + 4) + 7

Simplifying the equation:

3n + 6 = 2n + 11

Move all the terms with n to one side:

3n - 2n = 11 - 6

n = 5

So the first odd integer (n) is 5.

The second odd integer is n + 2 = 5 + 2 = 7.

The third odd integer is n + 4 = 5 + 4 = 9.

Therefore, the three consecutive odd integers that satisfy the given condition are 5, 7, and 9.

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