# What is the sum of the arithmetic sequence 174, 168, 162, ..., if there are 35 terms?

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### 1 Answer

The easiest way to look at this is to first find the end point for both series. Let's start with the first one.

We know that the series starts with 174, and clearly, we are subtracting 6 to get to each subsequent term. So, let's lay this out to see if we get a pattern:

`174, 174-6, 174-2*6, 174-3*6,...`

If you'll notice, each term is just 174 with some multiple of 6 subtracted. In the first term, we can treat it as `174-0*6`.

Based on this patter, we can come up with a formula that reflects that each term is just its one less than the term number (starting with 1) multiplied by six subtracted from 174:

`a_n = 174 - (n-1)*6`

So, to get our 35th term, we'll simply plug 35 into our formula for "n":

`a_35 = 174 - (35-1)*6 = -30`

So, our sum, S, becomes:

`S = 174 + 168 + 162 + ... + (-24) + (-30)`

Now, to solve this, it's easiest to apply the method Carl Gauss figured out in school when he was trying to solve the same problem. Let's add the sum S to itself, but let's reverse the terms:

`S = 174+168+162+...+(-24)+(-30)`

`S = (-30)+(-24)+...+162+168+174`

If you add the left sides and right sides, you get the following:

`2S = (174-30) + (168-24) + ... + (-30 + 174)`

Simplifying:

`2S = 144 + 144 + ... + 144`

Remember, we started with 35 terms, so we're going to be adding 144 to itself 35 times. In other words, adding the sums together the way we did gives us:

`2S = 35*144`

Which will be:

`2S = 5040`

Now, we can divide both sides by 2 to get our sum:

`S = 2520`

There you have it! Just find what the last couple of terms will be, add the sum to itself with the terms reversed, then divide by 2.

You can apply the same method to your other problem. I'll leave that to you to work out so you can get this down!

Hope that helps!

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