SOLUTION: WRITING RULES: Write the next term in the sequence. Then write a rule for the nth term. -5, 10, -15, 20, . . . (I understand that the sequence is multiples of 5, but I don't und

Algebra ->  Sequences-and-series -> SOLUTION: WRITING RULES: Write the next term in the sequence. Then write a rule for the nth term. -5, 10, -15, 20, . . . (I understand that the sequence is multiples of 5, but I don't und      Log On


   

Question 951201: WRITING RULES: Write the next term in the sequence. Then write a rule for the nth term.
-5, 10, -15, 20, . . .
(I understand that the sequence is multiples of 5, but I don't understand how they got the negative for every other number) THANKS!! :D :3
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The multiples of 5 can be represented by the expression 5n where n is a whole number

Ex: if n+=+2, then 5n+=+5%282%29+=+10

How do we alternate the signs? We introduce the term %28-1%29%5En. If n is even, then %28-1%29%5En=1 (ie positive 1). If n is odd, then %28-1%29%5En=-1. There are no other outcomes for this piece. So this is why %28-1%29%5En is the perfect thing to tack on to help alternate the signs.

The nth term is %28-1%29%5En%2A%285n%29. Notice how I am multiplying %28-1%29%5En by 5n

To confirm this, let's plug in n = 2 to get %28-1%29%5En%2A%285n%29+=+%28-1%29%5E2%2A%285%2A2%29+=+1%2A10+=+10.

Now let's plug in n = 3 to get %28-1%29%5En%2A%285n%29+=+%28-1%29%5E3%2A%285%2A3%29+=+-1%2A15+=+-15.

I'll let you check the others.

The next term is -25 as I'm sure you've already figured that out. Let's use n = 5 to confirm

%28-1%29%5En%2A%285n%29+=+%28-1%29%5E5%2A%285%2A5%29+=+-1%2A25+=+-25.
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