SOLUTION: The second term and the fifth term of a geometric series are 3 and 81 respectively. Find the sum from the fifth term to the tenth term of this series.

Algebra ->  Linear-equations -> SOLUTION: The second term and the fifth term of a geometric series are 3 and 81 respectively. Find the sum from the fifth term to the tenth term of this series.      Log On


   

Question 1070873: The second term and the fifth term of a geometric series are 3 and 81 respectively. Find the sum from the fifth term to the tenth term of this series.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
81=9%5E2=%283%5E2%29%5E2=3%5E%282%2A2%29=3%5E4
Mental math says that the common ratio is r=3 .
With formulas, id b%5B1%5D is the first term,
and r is the common ratio,
b%5Bn%5D=b%5B1%5D%2Ar%5E%28n-1%29 , so
b%5B5%5D=b%5B1%5D%2Ar%5E4 is 81 , the fifth term,
b%5B2%5D=b%5B1%5D%2Ar is 3 , the second term,
b%5B5%5D%2Fb%5B2%5D=b%5B1%5D%2Ar%5E4%2F%28b%5B1%5D%2Ar%29 ,
b%5B5%5D%2Fb%5B2%5D=r%5E3 , and substituting
81%2F3=r%5E3 ,
3%5E4%2F3=r%5E3 ,
3%5E3=r%5E3 , and r=3 .
The sum of n terms starting from a term b
in a geometric series with common ratio r , is
b%28r%5En-1%29%2F%28r-1%29 .
We are skipping the first 4 terms of a geometric series with r=3 ,
and adding up the next n=10-4=6 terms,
starting from fifth term b=81 and adding up to the tenth term,
so the sum is
81%283%5E6-1%29%2F%283-1%29=81%28729-1%29%2F2=81%2A728%2F2=81%2A364=highlight%2829484%29
You may think that I need to find the first term of that series,
then calculate the sum of the first 10 terms,
next calculate the sum of the first 4 terms,
and finally subtract one sum from the other.
That is inefficient.
Who can say that I am not allowed to think of my own series,
which happens to have the same r as the series in the problem,
and the same terms, except for missing the first 4?