Find the total number of integers between 10 and 1000 that are divisible by 3.

this simply becomes the arithmetic series

12 + 15 + 18 + .... + 999
where a = 12 , d = 3

so how many terms are there ?
termn = a + (n-1)d
999 = 12 + (n-1)(3)
987 = 3m - 3
990 = 3n
n =330

sum = (330/2)(24 + 329(3)) = 166815

or
sum = (n/2)(first + last)
= 165(12+999) = 166815

Thank u

To find the total number of integers between 10 and 1000 that are divisible by 3, we can use the concept of arithmetic sequences and the formula for the nth term of an arithmetic sequence.

An arithmetic sequence is a sequence of numbers in which the difference between any two consecutive terms is constant. In this case, the constant difference is 3 because we are looking for numbers that are divisible by 3.

The formula for the nth term of an arithmetic sequence is:
an = a1 + (n - 1)d

Where:
an is the nth term,
a1 is the first term,
n is the desired term number,
d is the common difference between the terms.

In our case, a1 = 12 (the smallest multiple of 3 greater than or equal to 10) and d = 3.

To find the number of terms in an arithmetic sequence, we can use the formula:
n = (l - a1) / d + 1

Where:
n is the number of terms,
a1 is the first term,
l is the last term,
d is the common difference.

In our case, a1 = 12 and l = 999 (the largest multiple of 3 less than or equal to 1000).

Now let's calculate the number of terms:

n = (999 - 12) / 3 + 1
n = 987 / 3 + 1
n = 329 + 1
n = 330

Therefore, there are 330 integers between 10 and 1000 that are divisible by 3.