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

$5050$

$4864$

$5544$

$4466$

✅ Correct Option: 3

We need to find the mean of all 4-digit even numbers of the form 'aabb' where a > 0.

The form 'aabb' means the first two digits are the same, and the last two digits are the same. For example: 1122, 3344, 5566, etc.

Since a > 0, the digit 'a' can be: 1, 2, 3, 4, 5, 6, 7, 8, 9 (9 choices)

Since the number must be even, the digit 'b' must be even: 0, 2, 4, 6, 8 (5 choices)

Total numbers = 9 choices for 'a' × 5 choices for 'b' = 45 numbers

Instead of listing all 45 numbers, we'll use algebra to find the sum quickly.

Any number of the form 'aabb' can be written as:

aabb = 1000a + 100a + 10b + b

= 1100a + 11b

Sum of all numbers = Σ(1100a + 11b) for all valid combinations

= 1100 × (sum of all 'a' values) + 11 × (sum of all 'b' values)

For 'a' values:

Each value (1, 2, 3, 4, 5, 6, 7, 8, 9) appears 5 times (once for each even 'b')

Sum of 'a' values = 5 × (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= 5 × 45 = 225

For 'b' values:

Each value (0, 2, 4, 6, 8) appears 9 times (once for each 'a')

Sum of 'b' values = 9 × (0 + 2 + 4 + 6 + 8)

= 9 × 20 = 180

Total sum = 1100 × 225 + 11 × 180

= 247500 + 1980 = 249480

Mean = $\tfrac{\text{Total sum}}{\text{Total count}} = \tfrac{249480}{45} = 5544$

Notice that 5544 is exactly the "middle" number when we arrange all possibilities systematically. This makes intuitive sense because the mean of a symmetric distribution equals the middle value.

Therefore, the mean is 5544.

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

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