What is the sum of the natural numbers from l to 100, which are not divisible by 7 and 11?

To solve this problem, we need to follow these steps:

1. First, calculate the sum of all natural numbers from 1 to 100. The formula to calculate the sum of first n natural numbers is n*(n+1)/2. So, the sum of first 100 natural numbers is 100*(100+1)/2 = 5050.

2. Then, calculate the sum of all natural numbers from 1 to 100 that are divisible by 7. The largest number less than 100 that is divisible by 7 is 98, and there are 98/7 = 14 such numbers. The sum of these numbers is 7*(14*(14+1)/2) = 735.

3. Similarly, calculate the sum of all natural numbers from 1 to 100 that are divisible by 11. The largest number less than 100 that is divisible by 11 is 99, and there are 99/11 = 9 such numbers. The sum of these numbers is 11*(9*(9+1)/2) = 550.

4. However, numbers that are divisible by both 7 and 11 have been subtracted twice. These are the numbers divisible by 77 (since 7*11 = 77). The largest number less than 100 that is divisible by 77 is 77 itself, and there is only one such number. So, add 77 back to the sum.

5. Finally, subtract the sums calculated in steps 2 and 3 from the sum calculated in step 1, and then add back the number calculated in step 4. This gives the sum of all natural numbers from 1 to 100 that are not divisible by 7 or 11.

So, the sum is 5050 - 735 - 550 + 77 = 3842.

Sure, let's solve this step by step.

Step 1: Find the sum of all natural numbers from 1 to 100. The formula to find the sum of first n natural numbers is n*(n+1)/2. So, the sum of first 100 natural numbers is 100*(100+1)/2 = 5050.

Step 2: Find the sum of all natural numbers from 1 to 100 that are divisible by 7. The numbers divisible by 7 between 1 and 100 are 7, 14, 21, ..., 98. This is an arithmetic progression with first term (a) as 7, last term (l) as 98 and common difference (d) as 7. The number of terms (n) can be calculated as n = (l - a)/d + 1 = (98 - 7)/7 + 1 = 14. The sum (S) of an arithmetic progression is n/2 * (a + l) = 14/2 * (7 + 98) = 735.

Step 3: Find the sum of all natural numbers from 1 to 100 that are divisible by 11. The numbers divisible by 11 between 1 and 100 are 11, 22, ..., 99. This is an arithmetic progression with first term (a) as 11, last term (l) as 99 and common difference (d) as 11. The number of terms (n) can be calculated as n = (l - a)/d + 1 = (99 - 11)/11 + 1 = 9. The sum (S) of an arithmetic progression is n/2 * (a + l) = 9/2 * (11 + 99) = 495.

Step 4: Find the sum of all natural numbers from 1 to 100 that are divisible by both 7 and 11. The numbers divisible by both 7 and 11 between 1 and 100 are 77. So, the sum is 77.

Step 5: Subtract the sums found in steps 2, 3 and 4 from the sum found in step 1. The sum of all natural numbers from 1 to 100 that are not divisible by 7 and 11 is 5050 - 735 - 495 + 77 = 3897.

We add 77 back because while subtracting the sums of numbers divisible by 7 and 11, we subtracted the sum of numbers divisible by 77 (which are divisible by both 7 and 11) twice. So, we add it back once to correct it.