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Find the sum of the geometric progression:

[tex]\[ 27 + 9 + 3 + \cdots + \frac{1}{9} \][/tex]


Sagot :

To find the sum of the geometric progression given by [tex]\(27, 9, 3, \cdots, \frac{1}{9}\)[/tex], we need to follow several steps. Here's a detailed, step-by-step solution:

1. Identify the first term ([tex]\(a\)[/tex]) and the common ratio ([tex]\(r\)[/tex]) of the series:

- The first term of the geometric progression ([tex]\(a\)[/tex]) is [tex]\(27\)[/tex].
- To find the common ratio ([tex]\(r\)[/tex]), we can divide the second term by the first term:
[tex]\[ r = \frac{9}{27} = \frac{1}{3} \][/tex]

2. Determine the number of terms ([tex]\(n\)[/tex]) in the progression:

- We observe the terms of the progression are [tex]\(27, 9, 3, 1, \frac{1}{3}, \frac{1}{9}\)[/tex].
- Counting these terms, we have [tex]\(6\)[/tex] terms in total. So, [tex]\(n = 6\)[/tex].

3. Use the formula for the sum of the first [tex]\(n\)[/tex] terms of a geometric series, [tex]\(S_n\)[/tex], given by:
[tex]\[ S_n = a \frac{1 - r^n}{1 - r} \][/tex]

4. Substitute the known values into the formula:
- [tex]\(a = 27\)[/tex]
- [tex]\(r = \frac{1}{3}\)[/tex]
- [tex]\(n = 6\)[/tex]

Therefore,
[tex]\[ S_6 = 27 \cdot \frac{1 - \left(\frac{1}{3}\right)^6}{1 - \frac{1}{3}} \][/tex]

5. Simplify the expression step-by-step:
- Calculate [tex]\(\left(\frac{1}{3}\right)^6\)[/tex]:
[tex]\[ \left(\frac{1}{3}\right)^6 = \frac{1}{729} \approx 0.001371742 \][/tex]
- Therefore,
[tex]\[ 1 - \frac{1}{729} \approx 1 - 0.001371742 = 0.998628258 \][/tex]

- Next, calculate [tex]\(1 - r\)[/tex]:
[tex]\[ 1 - \frac{1}{3} = \frac{2}{3} \][/tex]

- Plug these values back into the formula:
[tex]\[ S_6 = 27 \cdot \frac{0.998628258}{\frac{2}{3}} \][/tex]

- Simplify the fraction:
[tex]\[ \frac{0.998628258}{\frac{2}{3}} = 0.998628258 \times \frac{3}{2} = 1.497942387 \][/tex]

- Finally, the sum [tex]\(S_6\)[/tex] is:
[tex]\[ S_6 = 27 \cdot 1.497942387 \approx 40.44444444444444 \][/tex]

Thus, the sum of the geometric progression [tex]\(27 + 9 + 3 + \cdots + \frac{1}{9}\)[/tex] is approximately [tex]\(40.44\)[/tex].