How do you find #S_n# for the geometric series #a_2=-36#, a_5=972#, n=7?

Answer 1

#S_n=3^((n+1))(-1)^n-3#

The #n^(th)# term of a Geometric Series and its sum #S_n# up to n^(th)# term #a_n#, whose first term is #a_1# and common ratio is #r# is given by
#a_n=a_1xxr^((n-1))# and #S_n=a_1xx(r^n-1)/(r-1)#
as #a_2=a_1xxr=-36# and #a_5=a_1xxr^4=972#
Dividing latter by former, we get #r^3=-972/36=-27#
and hence #r=root(3)(-27)=-3#
and #a_1=36/-3=-12#
Hence, #S_n=-12xx((-3)^n-1)/((-3)-1)=-12xx((-3)^n-1)/(-4)# i.e.
#S_n=3((-3)^n-1)=3(3^n(-1)^n-1)=3^((n+1))(-1)^n-3#
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Answer 2

To find (S_n) for a geometric series, you can use the formula:

[ S_n = a_1 \frac{{1 - r^n}}{{1 - r}} ]

Where:

  • (S_n) is the sum of the first (n) terms,
  • (a_1) is the first term of the series,
  • (r) is the common ratio of the series, and
  • (n) is the number of terms.

Given (a_2 = -36) and (a_5 = 972), you can find the first term ((a_1)) and the common ratio ((r)). Then, substitute these values along with (n = 7) into the formula to find (S_7).

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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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