SOLUTION: Please view the document for the two questions.
https://docs.google.com/document/d/1Cl3HrcoR6jibNzxfDTIilS3imSplLY8nQUU0OpU9vLA/edit?usp=sharing
Algebra ->
Sequences-and-series
-> SOLUTION: Please view the document for the two questions.
https://docs.google.com/document/d/1Cl3HrcoR6jibNzxfDTIilS3imSplLY8nQUU0OpU9vLA/edit?usp=sharing
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Question 1181531: Please view the document for the two questions.
https://docs.google.com/document/d/1Cl3HrcoR6jibNzxfDTIilS3imSplLY8nQUU0OpU9vLA/edit?usp=sharing Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! a1= 120 first
common ratio = 1/5 ( ratio of consecutive terms)
last term = 24/78125
an = a1*r^(n-1)
24/78125 = 120*(1/5)^(n-1)
(1/5)^(n-1) = 1/390625
(n-1) log (1/5) = log(1/390625)
n-1=8
n=9 the number of terms
2-(4/3) +(8/9)-(16/27) ....
=> (2+(8/9) .....) - ((4/3) +16/27) ....)
Use the sum of n terms formula for both and simplify
