Proving the Difference of Sums in an Arithmetic Progression

[fluppocinonys]

Homework Statement

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 |.

Homework Equations

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

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?

 

[tiny-tim]

Hi fluppocinonys!

Do I use {u_n} as last term, and subsequently {u_{n - 1}}, {u_{n - 2}} for last second and third term?

Yes. ​

 

[fluppocinonys]

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

 

[tiny-tim]

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

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

 

[fluppocinonys]

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

 

[tiny-tim]

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

Yup!

and then you … ? ​