Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.

Answer

To find the sum of all natural numbers between 250 and 1000 that are divisible by 9, we can follow these steps:

### Step 1: Identify the first and last terms
First, we need to find the smallest natural number greater than 250 that is divisible by 9. We can do this by finding the smallest multiple of 9 that is greater than or equal to 250.

To find this, we divide 250 by 9:
\[ \frac{250}{9} \approx 27.78 \]
The smallest integer greater than 27.78 is 28. Therefore, the first term \( a \) is:
\[ a = 9 \times 28 = 252 \]

Next, we find the largest natural number less than 1000 that is divisible by 9. We divide 1000 by 9:
\[ \frac{1000}{9} \approx 111.11 \]
The largest integer less than or equal to 111.11 is 111. Therefore, the last term \( l \) is:
\[ l = 9 \times 111 = 999 \]

### Step 2: Determine the common difference
Since we are looking for numbers divisible by 9, the common difference \( d \) is:
\[ d = 9 \]

### Step 3: Find the number of terms
To find the number of terms \( n \) in this arithmetic progression (AP), we can use the formula for the \( n \)-th term of an AP:
\[ l = a + (n-1)d \]
Substituting the known values:
\[ 999 = 252 + (n-1) \times 9 \]
Now, we solve for \( n \):
\[ 999 - 252 = (n-1) \times 9 \]
\[ 747 = (n-1) \times 9 \]
\[ n-1 = \frac{747}{9} = 83 \]
\[ n = 84 \]

### Step 4: Calculate the sum of the terms
The sum \( S_n \) of the first \( n \) terms of an AP is given by the formula:
\[ S_n = \frac{n}{2} \times (a + l) \]
Substituting the values we have:
\[ S_{84} = \frac{84}{2} \times (252 + 999) \]
\[ S_{84} = 42 \times 1251 \]
Calculating this:
\[ S_{84} = 42 \times 1251 = 52542 \]

### Final Answer
The sum of all natural numbers between 250 and 1000 that are divisible by 9 is **52542**.

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