# Introductory Finite Mathematics problem..

• October 9th 2009, 12:38 PM
hippieguy
Introductory Finite Mathematics problem..
I'm not exactly sure where this question fits so I decided to post it here after trying several forums with no luck in receiving any responses. Hopefully someone here will be able to help me out since so many other forums have fallen short of my expectations. (Happy)

I'm taking a mandatory math class for my arts degree and have never been great, or good, at math. I'm working on my assignment that is due on Tuesday and have come to a stall with this question:

Find the sum of:
(1000-200) + (995-198) + (990-196) + ... + (505-2)

To solve it I am using these formulas:

n= An-A1/d+1

sum=A1+An/d *n

I have found that the difference is -3 and calculated that n must equal 100.

Now when I enter these terms into the formula that produces the sum I am given this: -15966.66667 which can't possibly be the correct answer.

Does anyone know where I went wrong? I am sure there are plenty of people here who would find this type of math very easy, it's a Math 101 course called Introductory to Finite Mathematics.

• October 9th 2009, 01:40 PM
CaptainBlack
Quote:

Originally Posted by hippieguy
I'm not exactly sure where this question fits so I decided to post it here after trying several forums with no luck in receiving any responses. Hopefully someone here will be able to help me out since so many other forums have fallen short of my expectations. (Happy)

I'm taking a mandatory math class for my arts degree and have never been great, or good, at math. I'm working on my assignment that is due on Tuesday and have come to a stall with this question:

Find the sum of:
(1000-200) + (995-198) + (990-196) + ... + (505-2)

To solve it I am using these formulas:

n= An-A1/d+1

sum=A1+An/d *n

Why, what are you trying to do?

CB
• October 9th 2009, 03:12 PM
Plato
Quote:

Originally Posted by hippieguy
Find the sum of:
(1000-200) + (995-198) + (990-196) + ... + (505-2)

I realize that this may be of little use to you.
But I recognize this a familiar sum $\sum\limits_{k = 1}^{100} {\left( {500 + 3k} \right)} = \left( {100} \right)\left( {500} \right) + 3\frac{{(100)(101)}}{2}$.

Maybe you can find something your textbook that looks like this.
If not just ignore it. But at least you know the correct answer.