# Thread: Find the sum of the integers...

1. ## Find the sum of the integers...

Hi

My problem is:

Find the sum of the integers between 200 and 499 that are multiples of three.

34,950

but I was wondering if there was an easier way to get to that besides adding all of the multiples of three? I kept getting messed up that way. Thank you.

2. Originally Posted by Averee
Hi

My problem is:

Find the sum of the integers between 200 and 499 that are multiples of three.
.
List them,

201,204,207,...,498

This is arithmetic sequence we commond difference 3.
Thus,
a_n=3n+b
Where, n is the "n-th" term.

Thus,
a_1=3(1)+b
But,
a_1=201
Thus,
201=3+b
And hence,
b=198

In general this sequence can be modelled as,
a_n=3n+198

Let us find what "n" is in the last term,
498=3n+198
300=3n
n=100

Thus, there are 100 terms.
Hence,
a_n=3n+198 where "n" ranges from 1 to 100.

We need to find the sum,
a_1+a_2+...+a_100
That is,
3(1)+198+3(2)+198+3(3)+198+...+3(100)+198
Rewrite this as,
3(1+2+...+100)+(198+198+...+198)
Where 198 appears 100 terms.
Thus,
3(1+2+...+100)+100*198

Now use the formula,
1+2+...+n=n(n+1)/2

Hence,

3(5050)+100*198

3. Thank you. I think I understand most of what you did, except your answer :
3(5050) + 100 * 198 equals 3,019,500, and my answer is supposed to be 34,950?

4. Originally Posted by Averee
Thank you. I think I understand most of what you did, except your answer :
3(5050) + 100 * 198 equals 3,019,500, and my answer is supposed to be 34,950?
My answer is correct and it matches with thine.

5. Okay, what did I do wrong?

6. Originally Posted by Averee
Okay, what did I do wrong?
I think you though that * was a comma. It means times.

100*198=19800

3(5050)=15150

15150+19800=34950

7. I see, I was doing the order of operations wrong. Thanks again. One more question though (promise), how did you get to that very last part? The one that gave the answer?

8. Originally Posted by Averee
I see, I was doing the order of operations wrong. Thanks again. One more question though (promise), how did you get to that very last part? The one that gave the answer?
What part.

9. I think I got most of it, but when you put 3(5050) + 100 * 198, how did you know to do that? Did you use one of the formulas you gave me?

10. Originally Posted by Averee
I think I got most of it, but when you put 3(5050) + 100 * 198, how did you know to do that? Did you use one of the formulas you gave me?
Yes.

1+2+3+...+100 = 101(100)/2=5050

By using the formula,

1+2+...+n=(n+1)n/2

11. Thanks again.