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Sagot :
To evaluate the infinite series using the formula provided, we need to identify the first term ([tex]\(a_1\)[/tex]) and the common ratio ([tex]\(r\)[/tex]) of the series.
The given series starts with [tex]\(12, 12, 2, \ldots\)[/tex].
1. First Term ([tex]\(a_1\)[/tex]):
The first term, denoted as [tex]\(a_1\)[/tex], can be directly seen from the series.
[tex]\[ a_1 = 12 \][/tex]
2. Common Ratio ([tex]\(r\)[/tex]):
The common ratio [tex]\(r\)[/tex] is found by dividing the second term by the first term. However, there seems to be an inconsistency in the series given because normally a geometric series should have a constant common ratio between consecutive terms. Here, the first two terms are 12 and 12, and the next term drops to 2.
To evaluate the situation:
- Calculation of common ratio between the first and second terms:
[tex]\[ r_1 = \frac{12}{12} = 1 \][/tex]
- Calculation of common ratio between the second and third terms:
[tex]\[ r_2 = \frac{2}{12} = \frac{1}{6} \][/tex]
Since the common ratio is not consistent, we should clarify whether the seriess in question is indeed geometric. Based on the given formula [tex]\(S = \frac{a_1}{1-r}\)[/tex], we typically assume a geometric progression with a common ratio [tex]\(r\)[/tex]. A converging infinite series requires that the common ratio [tex]\(r\)[/tex] satisfy [tex]\(|r| < 1\)[/tex].
Let's assume the given series is intended to be:
[tex]\[ 12, \left(12 \cdot \frac{1}{6}\right), \left(12 \cdot \left(\frac{1}{6}\right)^2\right), \ldots \][/tex]
3. Evaluating the Sum [tex]\(S\)[/tex]:
From the assumption, let’s use the more correct value of [tex]\(r = \frac{1}{6}\)[/tex]:
[tex]\[ r = \frac{1}{6} \][/tex]
Using the formula for the sum of an infinite geometric series [tex]\(S\)[/tex]:
[tex]\[ S = \frac{a_1}{1 - r} \][/tex]
Substituting the values:
[tex]\[ S = \frac{12}{1 - \frac{1}{6}} \][/tex]
Simplify the denominator:
[tex]\[ 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6} \][/tex]
Therefore:
[tex]\[ S = \frac{12}{\frac{5}{6}} = 12 \times \frac{6}{5} = 12 \times 1.2 = 14.4 \][/tex]
The evaluated sum, rounded to the nearest hundredth if necessary, is:
[tex]\[ \boxed{14.4} \][/tex]
The given series starts with [tex]\(12, 12, 2, \ldots\)[/tex].
1. First Term ([tex]\(a_1\)[/tex]):
The first term, denoted as [tex]\(a_1\)[/tex], can be directly seen from the series.
[tex]\[ a_1 = 12 \][/tex]
2. Common Ratio ([tex]\(r\)[/tex]):
The common ratio [tex]\(r\)[/tex] is found by dividing the second term by the first term. However, there seems to be an inconsistency in the series given because normally a geometric series should have a constant common ratio between consecutive terms. Here, the first two terms are 12 and 12, and the next term drops to 2.
To evaluate the situation:
- Calculation of common ratio between the first and second terms:
[tex]\[ r_1 = \frac{12}{12} = 1 \][/tex]
- Calculation of common ratio between the second and third terms:
[tex]\[ r_2 = \frac{2}{12} = \frac{1}{6} \][/tex]
Since the common ratio is not consistent, we should clarify whether the seriess in question is indeed geometric. Based on the given formula [tex]\(S = \frac{a_1}{1-r}\)[/tex], we typically assume a geometric progression with a common ratio [tex]\(r\)[/tex]. A converging infinite series requires that the common ratio [tex]\(r\)[/tex] satisfy [tex]\(|r| < 1\)[/tex].
Let's assume the given series is intended to be:
[tex]\[ 12, \left(12 \cdot \frac{1}{6}\right), \left(12 \cdot \left(\frac{1}{6}\right)^2\right), \ldots \][/tex]
3. Evaluating the Sum [tex]\(S\)[/tex]:
From the assumption, let’s use the more correct value of [tex]\(r = \frac{1}{6}\)[/tex]:
[tex]\[ r = \frac{1}{6} \][/tex]
Using the formula for the sum of an infinite geometric series [tex]\(S\)[/tex]:
[tex]\[ S = \frac{a_1}{1 - r} \][/tex]
Substituting the values:
[tex]\[ S = \frac{12}{1 - \frac{1}{6}} \][/tex]
Simplify the denominator:
[tex]\[ 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6} \][/tex]
Therefore:
[tex]\[ S = \frac{12}{\frac{5}{6}} = 12 \times \frac{6}{5} = 12 \times 1.2 = 14.4 \][/tex]
The evaluated sum, rounded to the nearest hundredth if necessary, is:
[tex]\[ \boxed{14.4} \][/tex]
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