The sum of all two digit numbers is :

1. 9810

2. 9045

3. 4509

4. 4905

Two digit numbers : 10, 11, ……., 99.

The above list is an A.P. with first term (a) = 10 and common difference (d) = 11 - 10 = 1.

Last term = 99.

Let n^th term be the last term.

∴ a_n = a + (n - 1)d

⇒ 99 = 10 + (n - 1)1

⇒ 99 = 10 + n - 1

⇒ 99 = n + 9

⇒ n = 99 - 9 = 90.

By formula,

Sum of n terms = $\dfrac{n}{2}[2a + (n - 1)d]$

Sum of the above A.P.

$= \dfrac{90}{2}[2 \times 10 + (90 - 1)(1)] \\[1em] = 45[20 + 89] \\[1em] = 45 \times 109 \\[1em] = 4905.$

Hence, Option 4 is the correct option.