Find the sum of all integers between 1 and 1000 that are divisible by 3

Help. It's an AS Maths question that isn't explained in the book. How would I do it?

If you think about it integers divisible by 3 are all 3 apart (aka a "common difference" of three). The first such number is evidently 3 (aka "first term") and 999 is the last.

With a first term and common difference we have an arithmetic sequence: a = 3 and d = 3

To find out the value of n which corresponds to the last term use the formula to find n: Un = a+(n-1)d
999 = 3 + 3(n-1) = 3n --> n = 333

Therefore you need to find S(333) using the formula

Sorry, I made a mistake originally, adding the 501 twice. You have to subtract 501 from the final sum if you use this technique, because 1000 is not evenly divisible by 3.

= (3 + 999) + (6 + 996) + (9 + 993) + . . . + (501 + 501)
= 1002 + 1002 + 1002 + . . . + 1002 =
= 1002*(501/3)
= 1002*167
= 167334 - 501 = 166883

3+6+9+12+...+999 =
3*(1+2+3+4+...+333).

Two ways to solve:

1) The sum of integers 1 to n equals n*(n+1)/2. In our problem, n=333. Thus, the total sum is 3*333*334/2 = 166833.

2) Pair the terms inside the parentheses as (1+333) + (2+332) + (3+331) + ... + (166+168) + 167. Each pairing sums to 334, and there are 166 pairings, plus an extra 167 by itself at the end. Thus, the total sum is 3*(334*166 + 167) = 166833.

3 + 6 + ... + 999 =
333
∑ 3i =
i=1

..333
3∑ i = 3(333)(334)/2 = 3(333)(167) = 266833
..i=1

s=n(a+l)/2
n=333 , a=3, l=999
s=166833