Question

Find the sum of the first 19 terms of the following sequence:

-11, -8, -5 . . .

Answer 1

\[ \large \sum_{n=0}^{18}(-11+3n)\]

Answer 2

what??

Answer 3

You have a sequence where the common difference, \(d\), is 3, it's an arithmetic sequence.

So, as a series, you can write it as:
\(a + (a+d) + (a+2d)+(a+3d) + ... +(a+18d)\) where "a + 18d" is the 19th term

Answer 4

The first term is \(a\), and you know that's -11. So you can write the sum as
\( -11+(-11+3)+(-11+2\cdot 3)+(-11+3 \cdot 3)+...+(-11+18\cdot 3)\)

Answer 5

Then you can try and see how you can use sigma notation to represent that sum. (What I first wrote)