The sum of the first ten terms of the sequence is -59,048. The correct option is C.-59,048
Here is the correct question:
What is the sum of the first ten terms of the sequence 4, -12, 36, -108...
From the question,
The given sequence is 4,-12, 36, -108...
This is a geometric progression since it has a common ratio.
The common ratio, r = -3
Recall that,
Given a geometric progression, the common ratio, r is given as
[tex]r = \frac{T_{2} }{T_{1}} = \frac{T_{3} }{T_{2}} =\frac{T_{4} }{T_{3}} = ...[/tex]
From the given sequence,
[tex]T_{1} = 4[/tex]
[tex]T_{2} = -12[/tex]
[tex]T_{3} = 36[/tex]
[tex]T_{4} = -108[/tex]
∴ [tex]r = \frac{-12 }{4} = \frac{36}{-12} =\frac{-108} {36} = -3[/tex]
Therefore, the sequence is a geometric progression with a common ratio of 3
Now, to determine the sum of the first ten terms of the sequence,
From the formula
[tex]S_{n} = \frac{a(1-r^{n}) }{1-r}[/tex]
Where [tex]S_{n}[/tex] is the sum of n terms
[tex]a[/tex] is the first term
[tex]r[/tex] is the common ratio
and [tex]n[/tex] is the number of terms in the series
From the question
[tex]a = 4[/tex]
[tex]r = -3[/tex]
and [tex]n = 10[/tex] (Since, we are to determine the sum of the first 10 terms)
Now, put the parameters into the formula
[tex]S_{n} = \frac{a(1-r^{n}) }{1-r}[/tex]
We get
[tex]S_{10} = \frac{4(1-(-3)^{10}) }{1-(-3)}[/tex]
[tex]S_{10} = \frac{4(1-59049) }{1+3}[/tex]
[tex]S_{10} = \frac{4(-59048) }{4}[/tex]
[tex]S_{10} = \frac{-236192 }{4}[/tex]
[tex]S_{10} = -59048[/tex]
Hence, the sum of the first ten terms of the sequence is -59,048. The correct option is C.-59,048
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