Question

What is the sum of a 36-term arithmetic sequence where the first term is 11 and the last term is 151?

Answer 1

so 36 terms are being added.. are they evenly spaced apart? you would need to know that.

Answer 2

yes

Answer 3

Use this formula

Sn=n/2 * (a+l) where n is the number of terms
a is the first term and l is the last term

Answer 4

@experimentX i dont think that gives you the sum. I tried it and got .1111 , im trying to think of a new formula.

Answer 5

@Jreis this should work out
http://upload.wikimedia.org/wikipedia/en/math/a/f/e/afe20f89d7bfdbd0a191168d80eb8077.png

Answer 6

Brilliant. 36/2 * (11+151) = 2916. Is that the right answer cheeto?

Answer 7

Jreis - The answer to that is yes, by definition of an arithmetic series.

Think about it this way: let the spacing between each number be \(n\), which will be consistent among the entire series. Now, consider writing the series in the two following ways, which both sum up to the same number, which we will call \(S\).
\[\begin{align}
11 + (11+n) + (11+2n) + \cdots + 115 &= S\\
115 + (115-n) + (115-2n) + \cdots + 11 &= S
\end{align}\]Now, let us add these two equations together, grouping together the first terms of each, the second terms of each, and so on.
\[(11+115)+(11+115+n-n) + (11+115+2n-2n) + \cdots (115+11) = 2S\]\[(11+115)+(11+115)+(11+115)+\cdots + (11+115)=2S\]
Since we are given that there are 36 terms, we can simply express this sum as \(36(11+115)\). This allows us to solve for \(S\). \[36(11+115)=2S\]\[S = \frac{36(11+115)}{2}=\boxed{2268}\]

Answer 8

Yeah, I understand it. It's 151 not 115 though, so its 2916. Good explanation though (but not my question :P)

Answer 9

Oops...well the same general scheme will still work, so I'll leave that as an exercise for the original poster. :)