SOLUTION: 1. In geometric sequence,the 2nd term is 2 and the 6th term is 2/81.Find the common ratio. 2.The sum of the first 5 terms of a geometric sequence is 93.If the 10th term is 3/2,fin

Algebra ->  Sequences-and-series -> SOLUTION: 1. In geometric sequence,the 2nd term is 2 and the 6th term is 2/81.Find the common ratio. 2.The sum of the first 5 terms of a geometric sequence is 93.If the 10th term is 3/2,fin      Log On


   

Question 1042111: 1. In geometric sequence,the 2nd term is 2 and the 6th term is 2/81.Find the common ratio.
2.The sum of the first 5 terms of a geometric sequence is 93.If the 10th term is 3/2,find the common ratio and the 4th term of the sequence
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
In a geometric sequence:
b%5B1%5D= the first term
r= the common ratio.
Any other term or the sum of terms starting from b%5B1%5D can be found from
b%5Bn%5D=b%5B1%5D%2Ar%5E%28%28n-1%29%29= the n%5Eth term, and
S%5Bn%5D=b%5B1%5D%2A%28%28r%5En-1%29%2F%28r-1%29%29= the sum of the first n terms.

1) The problem states that
b%5B2%5D=b%5B1%5D%2Ar%5E%28%282-1%29%29=b%5B1%5D%2Ar=2 and b%5B6%5D=b%5B1%5D%2Ar%5E%28%286-1%29%29=b%5B1%5D%2Ar%5E5=2%2F81 .
Was there any other information? Did it state that all the terms in the sequence were positive real numbers?
As posted, the problem has two solutions:
system%28b%5B1%5D%2Ar=2%2Cb%5B1%5D%2Ar%5E5=2%2F81%29--->b%5B1%5D%2Ar%5E5%2F%28b%5B1%5D%2Ar%29=%28%282%2F81%29%29%2F2--->r%5E4=1%2F81 .
There are two real number solutions to r%5E4=81 :
either r=-1%2F3 or r=3 .
If r=-3 , the sequence first term is -6 , and the next five terms are .
If r=3 , the sequence first term is 6 , and the next five terms are .

2) The problem (as posted) states that
b%5B10%5D=b%5B1%5D%2Ar%5E%28%2810-1%29%29=b%5B1%5D%2Ar%5E9=3%2F2 and S%5B5%5D=b%5B1%5D%2A%28%28r%5E5-1%29%2F%28r-1%29%29=93 .
From there, we get
system%28b%5B1%5D%2Ar%5E9=3%2F2%2Cb%5B1%5D%2A%28%28r%5E5-1%29%2F%28r-1%29%29=93%29--->b%5B1%5D%2A%28%28r%5E5-1%29%2F%28r-1%29%29%2F%28b%5B1%5D%2Ar%5E9%29=93%2F%28%283%2F2%29%29--->>%28r%5E5-1%29%2F%28%28r-1%29%2Ar%5E9%29=93%2A%282%2F3%29--->%28r%5E5-1%29%2F%28%28r-1%29%2Ar%5E9%29=62--->%28r%5E5-1%29%2F%28r-1%29=62%2Ar%5E9--->r%5E4%2Br%5E3%2Br%5E2%2Br%2B1=62%2Ar%5E9--->62%2Ar%5E9-r%5E4-r%5E3-r%5E2-r-1=0
That last equation has no rational solution. It has only one real, irrational solution that can be approximated by r=0.7094609 .
It is easy to find such an approximate solution if you have a graphing calculator or the right computer software.
With that, we can calculate b%5B1%5D from
b%5B1%5D%2Ar%5E9=3%2F2 :
b%5B1%5D%2A0.7094609%5E9=1.5--->b%5B1%5D=1.5%2F0.7094609%5E9--->b%5B1%5D=32.94087 (also an approximation), and
b%5B4%5D=b%5B1%5D%2Ar%5E%28%284-1%29%29=b%5B1%5D%2Ar%5E3=32.94087%2A0.7094609%5E3=11.76306 (approximately).
Did your instructor really epect you to solve the problem that way? Or is there a typo somewhere, so that there would be an easier, rational solution for r ?