The mean of all 4-digit even natural numbers of the form 'aabb', where a > 0, is

A 4-digit number of the form 'aabb' is written as:
aabb = 1000a + 100a + 10b + b = 1100a + 11b

For the number to be even, the last digit 'b' must be even. So, b can take values from {0, 2, 4, 6, 8}.
Since it is a 4-digit number, a > 0. Thus, a can take values from {1, 2, 3, 4, 5, 6, 7, 8, 9}.

For each value of a, there are 5 possible values of b. Therefore, there are 9 × 5 = 45 such numbers.

The sum of all these numbers is:
Sum = Σ (1100a + 11b) = 5 × 1100 × Σ_a=1⁹ a + 9 × 11 × Σ_{b ∈ {0,2,4,6,8}} b
Sum = 5500 × (1 + 2 +... + 9) + 99 × (0 + 2 + 4 + 6 + 8)
Sum = 5500 × 45 + 99 × 20
Sum = 247500 + 1980 = 249480

The mean of these numbers is:
$\text{Mean}=\frac{249480}{45}=5544$