# Simple Series Question.

1. Mar 29, 2005

### misogynisticfeminist

Evaluate the sum of all integers from 4 to 1000 (4 and 1000 included) excluding all multiples of both 3 and 7.

What i did was,

$$\sum_{r=4}^{996} r=\sum_{r=1}^{1000} r - \sum_{r=1}^{4} r$$

Well, the hard part was excluding multiples of both 3 and 7. The sequence of these numbers would be,

$$21, 42, 63,...n$$

But however, i find no common ratio even between the 1st 3 terms. So, it is not a geometric progression. How do I evaluate a series of terms whose multiples are both 3 and 7 if they are not APs or GPs?

Thanks alot.

2. Mar 29, 2005

### SpaceTiger

Staff Emeritus
Can't you get that by subtracting off multiples of smaller arithmetic progressions? For example, the sum from 1 to 100, excluding multiples of 5 is:

$$\sum_{n=1}^{100}r-5\sum_{n=1}^{20}r$$

3. Mar 29, 2005

### SpaceTiger

Staff Emeritus
Sorry, I just realized that I skimmed your question too quickly. To deal with common multiples, consider the fact that if you do what I just said for both 3 and 7, you would subtract the common multiples twice. You don't want this, but if you then add those back in once (that is, add back in the series you listed), you'll be alright. And if you don't think that series is an AP, check again.

Last edited: Mar 29, 2005
4. Mar 29, 2005

### misogynisticfeminist

hey, i just realized that 21,42,63,... is an AP, thanks alot.