SOLUTION: did I do this right: Use the geometric sequence of numbers 1, 3, 9, 27, … to find the following: a) What is r, the ratio between 2 consecutive terms? Answer: r=3 Show work in

Algebra ->  Sequences-and-series -> SOLUTION: did I do this right: Use the geometric sequence of numbers 1, 3, 9, 27, … to find the following: a) What is r, the ratio between 2 consecutive terms? Answer: r=3 Show work in       Log On


   



Question 39248: did I do this right: Use the geometric sequence of numbers 1, 3, 9, 27, … to find the following:
a) What is r, the ratio between 2 consecutive terms?
Answer: r=3
Show work in this space.
each are multiples of 3
1*3=3, 3*3=9, 9*3=27 and so on

b) Using the formula for the nth term of a geometric sequence, what is the 10th term?
Answer:n=19683
Show work in this space.
a(n)=a(1)(r^n-1)
a(n)=1(1)*(3^10-1
a(n)=3^9=19683
c) Using the formula for the sum of a geometric series, what is the sum of the first 10 terms?
Answer: s=29524
Show work in this space.
s(n)= a(1)(1-r^n)/1-r
s(n)= 1(1)*(1-3^10)/(1-3)=29524

Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
It all looks good. The only thing I would mention is that the way to find the common ratio of a geometric sequence or series is to merely divide any a-sub-(n+1) term by a-sub-n...