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what is the general term of the following series? 60/121-30/11+15-....+219615/16?
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High School Teacher
I answered this in the previous question.
First we need to find out what `(60/121)*x = -30/11` and we get
`x = -30/11*121/60 = -11/2`
We can see that `+15 = (-30/11) * (-11/2) = 15` so this sequence seems to be the previous term multiplied by `(-11/2).`
So we can write this as `A_1(-11/2)^(n-1)` where `A_1 = 60/121` in this case
So the general term of this sequence is `60/121(-11/2)^(n-1)` or we could write it as
`60/121(-11/2)^h*(-2/11) = -120/1331(-11/2)^n`
Posted by beckden on March 11, 2012 at 9:17 AM (Answer #1)
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