At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Given a geometric series with the first term [tex]\(a_T = 10\)[/tex] and common ratio [tex]\(r = \frac{1}{5}\)[/tex], we need to find the sum of the first five terms, expressed as an improper fraction in lowest terms.
To find the sum of the first [tex]\(n\)[/tex] terms of a geometric series, we use the formula:
[tex]\[ S_n = a_T \frac{1 - r^n}{1 - r} \][/tex]
For our specific problem:
- The first term [tex]\(a_T = 10\)[/tex]
- The common ratio [tex]\(r = \frac{1}{5}\)[/tex]
- The number of terms [tex]\(n = 5\)[/tex]
First, calculate [tex]\(r^5\)[/tex]:
[tex]\[ r^5 = \left(\frac{1}{5}\right)^5 = \frac{1}{3125} \][/tex]
Next, compute [tex]\(1 - r^5\)[/tex]:
[tex]\[ 1 - r^5 = 1 - \frac{1}{3125} = \frac{3125}{3125} - \frac{1}{3125} = \frac{3124}{3125} \][/tex]
Now, use the sum formula:
[tex]\[ S_5 = 10 \cdot \frac{\frac{3124}{3125}}{1 - \frac{1}{5}} \][/tex]
Simplify the denominator [tex]\(1 - \frac{1}{5}\)[/tex]:
[tex]\[ 1 - \frac{1}{5} = \frac{5}{5} - \frac{1}{5} = \frac{4}{5} \][/tex]
Substitute back into the sum formula:
[tex]\[ S_5 = 10 \cdot \frac{\frac{3124}{3125}}{\frac{4}{5}} \][/tex]
Simplify the fraction inside the sum formula:
[tex]\[ S_5 = 10 \cdot \left( \frac{3124}{3125} \cdot \frac{5}{4} \right) \][/tex]
Multiply the fractions:
[tex]\[ S_5 = 10 \cdot \frac{3124 \cdot 5}{3125 \cdot 4} = 10 \cdot \frac{15620}{12500} = 10 \cdot \frac{1562}{1250} \][/tex]
Now, simplify to lowest terms:
[tex]\[ \frac{1562}{1250} = \frac{781}{625} \][/tex]
Thus, the sum of the first five terms of the geometric series is:
[tex]\[ S_5 = 10 \cdot \frac{781}{625} = \frac{7810}{625} \][/tex]
Finally, express the improper fraction in lowest terms:
[tex]\[ \frac{7810}{625} \][/tex]
Therefore, the sum of the first five terms of the geometric series is:
[tex]\[ \frac{7810}{625} \][/tex]
Answer here:
[tex]\[ \frac{7810}{625} \][/tex]
To find the sum of the first [tex]\(n\)[/tex] terms of a geometric series, we use the formula:
[tex]\[ S_n = a_T \frac{1 - r^n}{1 - r} \][/tex]
For our specific problem:
- The first term [tex]\(a_T = 10\)[/tex]
- The common ratio [tex]\(r = \frac{1}{5}\)[/tex]
- The number of terms [tex]\(n = 5\)[/tex]
First, calculate [tex]\(r^5\)[/tex]:
[tex]\[ r^5 = \left(\frac{1}{5}\right)^5 = \frac{1}{3125} \][/tex]
Next, compute [tex]\(1 - r^5\)[/tex]:
[tex]\[ 1 - r^5 = 1 - \frac{1}{3125} = \frac{3125}{3125} - \frac{1}{3125} = \frac{3124}{3125} \][/tex]
Now, use the sum formula:
[tex]\[ S_5 = 10 \cdot \frac{\frac{3124}{3125}}{1 - \frac{1}{5}} \][/tex]
Simplify the denominator [tex]\(1 - \frac{1}{5}\)[/tex]:
[tex]\[ 1 - \frac{1}{5} = \frac{5}{5} - \frac{1}{5} = \frac{4}{5} \][/tex]
Substitute back into the sum formula:
[tex]\[ S_5 = 10 \cdot \frac{\frac{3124}{3125}}{\frac{4}{5}} \][/tex]
Simplify the fraction inside the sum formula:
[tex]\[ S_5 = 10 \cdot \left( \frac{3124}{3125} \cdot \frac{5}{4} \right) \][/tex]
Multiply the fractions:
[tex]\[ S_5 = 10 \cdot \frac{3124 \cdot 5}{3125 \cdot 4} = 10 \cdot \frac{15620}{12500} = 10 \cdot \frac{1562}{1250} \][/tex]
Now, simplify to lowest terms:
[tex]\[ \frac{1562}{1250} = \frac{781}{625} \][/tex]
Thus, the sum of the first five terms of the geometric series is:
[tex]\[ S_5 = 10 \cdot \frac{781}{625} = \frac{7810}{625} \][/tex]
Finally, express the improper fraction in lowest terms:
[tex]\[ \frac{7810}{625} \][/tex]
Therefore, the sum of the first five terms of the geometric series is:
[tex]\[ \frac{7810}{625} \][/tex]
Answer here:
[tex]\[ \frac{7810}{625} \][/tex]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.