Question

How do you find the sum of the first 10 terms of `6+3+0+(-3)+(-6)+...`?

Answer 1

`-75`

Explanation:

I'm going to derive the formula we need because it's always handy to know where things come from, and quite frankly it's fairly interesting too. If you just came for the answer skip to the end but hopefully you'll have a look at the derivation.

`color(blue)("Derivation Start")`

For an arithmetic sequence, the `n^(th)` term is given by :

`x_n = a + (n-1)d`

So, to sum up to the nth term we first write it in sigma notation:

`sum_(k=0)^(n-1) (a + kd)`

You can see that this will spit out:

`a + (a+d) + (a+2d) + ...(a+(n-2)d) + (a+(n-1)d)`

Which is the sum we are looking for. We shall call it `S_n`

I'm now going to derive the formula for this sum. This is quite a cheap trick, but as you progress through mathematics you'll often find it's easier to understand things and remember a few wee tricks than to try and remember oodles of formulae.

Take our sum:

`S_n = a + (a+d) + (a+2d) + ...+(a+(n-2)d) + (a+(n-1)d)`

Reverse it:

`S_n = (a+(n-1)d) +(a+(n-2)d) + ...+ (a+2d) + (a+d) + a`

Add these two expressions up, term by term to obtain `2S_n`. You'll notice that every pair of terms comes out the same!

`a + (a+(n-1)d) = 2a+(n-1)d`

`(a+d) + (a+(n-2)d) = 2a + (n-1)d`

And so on, in short, every term-pair sum comes out to be equal to:

`2a+(n-1)d` and there are `n` pairs, so:

`2S_n = n*(2a+(n-1)d)`

`therefore S_n = n/2(2a+(n-1)d)`

`color(blue)("Derivation End")`

Back to the question. We want the sum of the first 10 terms.

We have arithmetic sequence with first term `a= 6` and common difference `d = -3`.

`S_(10) = (10)/(2)(2(6) + (10-1)(-3))`

`=5(12 - 27) = -75`