Find the sum of all three-digit numbers divisible by 9 .

Correct option is C

Given:
Numbers are three-digit integers divisible by 9.
Formula Used:
Arithmetic Progression formulas:
Nth term = a + (n - 1)d
Sum of n terms = $$\frac{n}{2}$$​ × (a + l)
Solution:
The smallest three-digit number divisible by 9 is 108.
The largest three-digit number divisible by 9 is 999.
The numbers form an Arithmetic Progression (AP): 108, 117, 126, ..., 999.
Here, first term (a) = 108, last term (l) = 999, common difference (d) = 9.
Find the number of terms (n):
999 = 108 + (n - 1) × 9
891 = (n - 1) × 9
n - 1 = 99
n = 100
Now, find the sum of these 100 terms.
Sum = $$\frac{100}{2} × (108 + 999)$$​​
= 50 × 1107
= 55350
Final Answer
So the correct answer is (c)