Find all sets of three consecutive multiples of 11 for which the sum of two lesser numbers is greater than 100 and the sum of the two greater numbers is less than 200

11x + 11(x+1) > 100

11(x+1) + 11(x+2) < 200

22x > 89
22x < 167

4.04 < x < 7.59

So, the numbers can be

55 66 77
66 77 88
77 88 99

To solve this problem, we need to find three consecutive multiples of 11 that satisfy two conditions:

1. The sum of the two smaller numbers is greater than 100.
2. The sum of the two larger numbers is less than 200.

Let's break down the problem into steps:

Step 1: Find the multiples of 11.
We start by writing down the first few multiples of 11:
11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, ...

Step 2: Identify three consecutive multiples of 11.
We can start by taking the first three multiples of 11: 11, 22, 33.

Step 3: Check the conditions.
Condition 1: The sum of the two smaller numbers is greater than 100.
11 + 22 = 33 > 100. Satisfied!

Condition 2: The sum of the two larger numbers is less than 200.
22 + 33 = 55 < 200. Satisfied!

The set {11, 22, 33} is a valid solution.

Step 4: Continue finding other sets.
We can repeat steps 2 and 3 to find more sets of three consecutive multiples of 11 that satisfy the given conditions.

Taking the next three multiples of 11: 22, 33, 44.

Condition 1: 22 + 33 = 55 > 100. Satisfied!

Condition 2: 33 + 44 = 77 < 200. Satisfied!

The set {22, 33, 44} is another valid solution.

We can continue this process to find more sets of three consecutive multiples of 11 that meet the criteria.

However, it is not possible to explain all the sets here as there are infinitely many sets of three consecutive multiples of 11.