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To find the sum of the first ten terms in the given geometric series [tex]\(16, -48, 144, -432, \ldots\)[/tex], we'll follow these steps.
First, identify the initial term [tex]\(a\)[/tex] and the common ratio [tex]\(r\)[/tex] of the series:
- The initial term [tex]\(a\)[/tex] is the first term in the series, which is [tex]\(16\)[/tex].
- The common ratio [tex]\(r\)[/tex] can be found by dividing the second term by the first term, which gives [tex]\(r = \frac{-48}{16} = -3\)[/tex].
We need to find the sum of the first ten terms in this geometric series, denoted as [tex]\(S_{10}\)[/tex].
The formula for the sum [tex]\(S_n\)[/tex] of the first [tex]\(n\)[/tex] terms of a geometric series is:
[tex]\[ S_n = a \frac{1 - r^n}{1 - r} \quad \text{for} \quad r \neq 1 \][/tex]
Let's plug in the values:
- [tex]\(a = 16\)[/tex]
- [tex]\(r = -3\)[/tex]
- [tex]\(n = 10\)[/tex]
Substitute these values into the formula:
[tex]\[ S_{10} = 16 \frac{1 - (-3)^{10}}{1 - (-3)} \][/tex]
Calculate [tex]\((-3)^{10}\)[/tex]:
[tex]\[ (-3)^{10} = 59049 \][/tex]
Now, substitute back into the formula:
[tex]\[ S_{10} = 16 \frac{1 - 59049}{1 + 3} \][/tex]
Simplify the expression:
[tex]\[ S_{10} = 16 \frac{1 - 59049}{4} = 16 \frac{-59048}{4} \][/tex]
[tex]\[ S_{10} = 16 \times (-14762) \][/tex]
[tex]\[ S_{10} = -236192 \][/tex]
Thus, the sum of the first ten terms in the geometric series is [tex]\(-236,192\)[/tex].
The correct answer is:
[tex]\[ \boxed{-236192} \][/tex]
First, identify the initial term [tex]\(a\)[/tex] and the common ratio [tex]\(r\)[/tex] of the series:
- The initial term [tex]\(a\)[/tex] is the first term in the series, which is [tex]\(16\)[/tex].
- The common ratio [tex]\(r\)[/tex] can be found by dividing the second term by the first term, which gives [tex]\(r = \frac{-48}{16} = -3\)[/tex].
We need to find the sum of the first ten terms in this geometric series, denoted as [tex]\(S_{10}\)[/tex].
The formula for the sum [tex]\(S_n\)[/tex] of the first [tex]\(n\)[/tex] terms of a geometric series is:
[tex]\[ S_n = a \frac{1 - r^n}{1 - r} \quad \text{for} \quad r \neq 1 \][/tex]
Let's plug in the values:
- [tex]\(a = 16\)[/tex]
- [tex]\(r = -3\)[/tex]
- [tex]\(n = 10\)[/tex]
Substitute these values into the formula:
[tex]\[ S_{10} = 16 \frac{1 - (-3)^{10}}{1 - (-3)} \][/tex]
Calculate [tex]\((-3)^{10}\)[/tex]:
[tex]\[ (-3)^{10} = 59049 \][/tex]
Now, substitute back into the formula:
[tex]\[ S_{10} = 16 \frac{1 - 59049}{1 + 3} \][/tex]
Simplify the expression:
[tex]\[ S_{10} = 16 \frac{1 - 59049}{4} = 16 \frac{-59048}{4} \][/tex]
[tex]\[ S_{10} = 16 \times (-14762) \][/tex]
[tex]\[ S_{10} = -236192 \][/tex]
Thus, the sum of the first ten terms in the geometric series is [tex]\(-236,192\)[/tex].
The correct answer is:
[tex]\[ \boxed{-236192} \][/tex]
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