How do you find the number of terms given #s_n=-66.67# and #-90+30+(-10)+10/3+...#?

Answer 1

The sum of first #4# terms is #s_4=-66.67#

Let us denote, by #t_n# the #n^(th)# term of the given series.Then, we find that,#t_2/t_1=t_3/t_2=t_4/t_3=...=-1/3#.

We conclude that it is a Geometric Series, with, common ratio

#r=-1/3, and, t_1=-90#
The sum #s_n# of first #n# terms of the series is given by,
#s_n=(t_1(1-r^n))/(1-r)#
Our goal is to find #n#, given, #s_n=-66.67=-66 2/3=-200/3, &, r=-1/3#.
#:. -200/3=(-90){(1-(-1/3)^n)/(1+1/3)}#
#:. -200/3(1/-90)(4/3)=80/81={1-(-1/3)^n}#
#:. (-1/3)^n=1-80/81=1/81=(-1/3)^4#
#:. n=4#
Hence, the sum of first #4# terms is #s_4=-66.67#
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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