SOLUTION: Can ANYONE help me with these two problems? Please! Type the general term for the following sequences: #1) 3, 9, 27, 81, 243, . . . __________ #2) 3/4, 6/5, 9/6, 12

Algebra ->  Sequences-and-series -> SOLUTION: Can ANYONE help me with these two problems? Please! Type the general term for the following sequences: #1) 3, 9, 27, 81, 243, . . . __________ #2) 3/4, 6/5, 9/6, 12      Log On


   

Question 202544: Can ANYONE help me with these two problems? Please!

Type the general term for the following sequences:

#1) 3, 9, 27, 81, 243, . . .
__________

#2) 3/4, 6/5, 9/6, 12/7, 15/8...
__________

Thanks,
Sarah
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first one to get you going

# 1



Notice that each term is increasing exponentially. So this sequence might be a geometric sequence. To find out, let's simply divide the terms.


First divide the 2nd term 9 by the 1st term 3 to get
9%2F3=3

Now divide the 3rd term 27 by the 2nd term 9 to get
27%2F9=3

Now divide the 4th term 81 by the 3rd term 27 to get
81%2F27=3

Now divide the 5th term 243 by the 4th term 81 to get
243%2F81=3

So if we pick ANY term and divide it by the previous term, we'll always get 3.

This is the common ratio between the terms. So this means that r=3.



Now since we've started at 3, this means that a=3



Since the general geometric sequence is a%5Bn%5D=ar%5En, this means the sequence is

a%5Bn%5D=3%2A3%5En where n starts at n=0


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Answer:

So the sequence is a%5Bn%5D=3%2A3%5En where n starts at n=0

Note: if you want to start at n=1, then you need to replace "n" with "n-1" to get the new sequence a%5Bn%5D=3%2A3%5E%28n-1%29