# Arithmetic Progression

1. Mar 4, 2009

### fluppocinonys

1. The problem statement, all variables and given/known data
An arithmetic progression has n terms and a common difference of d. Prove that the difference between the sum of the last k terms and the sum of the first k terms is | (n-k)kd |.

2. Relevant equations

$$\begin{array}{l} {S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right] \\ {u_n} = {a_1} + \left( {n - 1} \right)d \\ \end{array}$$

3. The attempt at a solution
I have no idea how to apply the "first 3 terms" and "last 3 terms" into the equation...
Do I use $${u_n}$$ as last term, and subsequently $${u_{n - 1}}$$, $${u_{n - 2}}$$ for last second and third term?

2. Mar 4, 2009

### tiny-tim

Hi fluppocinonys!
Yes.

3. Mar 5, 2009

### fluppocinonys

I tried but still unable to solve it.
Can you please hint me on how to start the question with?
thanks

4. Mar 5, 2009

### tiny-tim

Hint: if n = 100, and k = 6, what is the difference between a1 and a95?

5. Mar 5, 2009

### fluppocinonys

a95 = a1 + 94d
so,
a95 - a1
= a1 + 94d - a1
= 94d

6. Mar 5, 2009

### tiny-tim

Yup!

and then you … ?