3 consecutive odd integers such that thrice the middle is 15 more than the sum of the other 2

3 consecutive odd integers such that thrice the middle is 15 more than the sum of the other 2.

• Let the first integer be n
• The next odd one (middle) is n + 2.
• The next odd one is n + 4

We are given 3(n + 2) = n + n + 4 + 15.

Simplifying, we get:
3n + 6 = 2n + 19

Plugging that problem into our search engine, we get n = 13.
So the next odd integer is 13 + 2 = 15
The next odd integer is 15 + 2 = 17