SOLUTION: the fourth term of a GP is 1/2 and the sixth term is 1/8. its common ratio is negative a) how many term of GP add up to give a sum of -2/22/32? b)what is the sum to infinity for

Algebra ->  Equations -> SOLUTION: the fourth term of a GP is 1/2 and the sixth term is 1/8. its common ratio is negative a) how many term of GP add up to give a sum of -2/22/32? b)what is the sum to infinity for      Log On


   

Question 1116337: the fourth term of a GP is 1/2 and the sixth term is 1/8. its common ratio is negative
a) how many term of GP add up to give a sum of -2/22/32?
b)what is the sum to infinity for this GP
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
ONE WAY TO START:
The recursive relation for a geometric progression,
relating term number n , a%5Bn%5D to the preceding term, a%5Bn-1%5D is
a%5Bn%5D=a%5Bn-1%5D%2Ar , where r is the common ratio.
So, applying that to the fourth, fifth and sixth terms
a%5B5%5D=a%5B4%5D%2Ar=%281%2F2%29%2Ar ,
a%5B6%5D=a%5B5%5D%2Ar=%28%281%2F2%29%2Ar%29%2Ar=%281%2F2%29r%5E2 , and
%281%2F2%29r%5E2=1%2F8 --> r%5E2=%281%2F8%29%2F%281%2F2%29 --> r%5E2=1%2F4 .
Knowing that r%3C0 , we conclude that r=-1%2F2 ,
and can write the first 4 terms, starting from the 4th,
and dividing by %28-1%2F2%29 (multiplying times -2)
each time to get the term before:
a%5B4%5D=1%2F2 ,
a%5B3%5D=%281%2F2%29%2F%28-1%2F2%29=1%2F2%29%28-2%29=-1 ,
a%5B2%5D=%28-1%29%28-2%29=2 ,
a%5B1%5D=1%28-2%29=-4 .

ANOTHER WAY:
The general "formula" for term number n , a%5Bn%5D ,
of a GP with first term a%5B1%5D and common ratio r is
a%5Bn%5D=a%5B1%5D%2Ar%5E%28n-1%29 .
Applying that to system%28a%5B4%5D=1%2F2%2Ca%5B6%5D=1%2F8%2Cr%3C0%29 ,
system%28a%5B1%5D%2Ar%5E3=1%2F2%2Ca%5B4%5D%2Ar%5E6=1%2F8%2Cr%3C0%29 --> system%28a%5B1%5D%2Ar%5E3=1%2F2%2Cr%5E2=%281%2F8%29%2F%281%2F2%29=1%2F4%2Cr%3C0%29 --> system%28a%5B1%5D%2A%28-1%2F2%29%5E3=1%2F2%2Cr=-1%2F2%29 --> system%28a%5B1%5D%2A%28-1%2F8%29=1%2F2%2Cr=-1%2F2%29 --> system%28a%5B1%5D=-4%2Cr=-1%2F2%29

AND FROM THERE:
The "formula" for the sum of the first n terms, S%5Bn%5D , of a GP is
S%5Bn%5D=a%5B1%5D%28r%5En-1%29%2F%28r-1%29
When r%3C1 ,
the term r%5En tends to 0 as n tends to infinity ,
and the sum to infinity is
S%5Binfinity%5D=%28-a%5B1%5D%29%2F%281-r%29 .
In this case the sum to infinity is easy to calculate:
S%5Binfinity%5D%22=%22%28-%28-4%29%29%2F%28%28-1%2F2-1%29%29%22=%224%2F%28%28-3%2F2%29%29%22=%224%28-2%2F3%29%22=%22highlight%28-8%2F3=-2%262%2F3=%22-2.6666...%22%29
Substituting the a%5B1%5D and r values found above,
the expression for the sum of the first n terms becomes
S%5Bn%5D%22=%22%28-%28-4%29%29%28%28-1%2F2%29%5En-1%29%2F%28-3%2F2%29%22=%224%28-2%2F3%29%28%28-1%2F2%29%5En-1%29%22=%22%288%2F3%29%28%28%28-1%29%5En-2%5En%29%2F2%5En%29%22=%22%282%5E3%2F3%29%28%28%28-1%29%5En-2%5En%29%2F2%5En%29%22=%22%281%2F2%5E%28n-3%29%29%28%28%28-1%29%5En-2%5En%29%2F3%29
If that is supposed to be an irreducible fraction of the form K%2F32 ,
the only possibility is 2%5E%28n-3%29=32 ,
and as 32=2%5E5 , n-3=5 --> highlight%28n=8%29 .

BOTH QUESTIONS WERE ANSWERED, BUT IF YOU WANT MORE,
knowing n , we could try to find that denominator.
.
So,
S%5Binfinity%5D%22=%22 .