# what is the general term of the following series? 60/121-30/11+15-....+219615/16?

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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`