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rosedewittbukater

  • 3 years ago

Can someone help me and explain this Algebra 2? 12. Describe an infinite geometric series with a beginning value of 2 that converges to 10. What are the first 4 terms of the series? (2 points)

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  1. rosedewittbukater
    • 3 years ago
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    I tried using 2 for a1 and 10 for s and found r as -4 but I couldn't get an equation to find the first four terms. And I don't know a way to find the terms that will add up to 10. Help anyone?

  2. campbell_st
    • 3 years ago
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    the limiting sum of a geometric series is used when a series converges... so the formula is \[S_{\infty} = \frac{a}{1 -r}\] a = 1st term and r is the common ratio. the value of r will be \[|r|<1\] this is what makes it converge In your question you will need to find r, the common ratio to allow you to find the 4 terms so using your information \[10 = \frac{2}{1 - r} \] \[1 - r = \frac{1}{5} \] which will result in \[r = \frac{4}{5} \] so to find the 2nd, 3rd and 4th terms use the formula \[T_{n} = ar^{n -1} \] you know the 1st term \[T_{2} = 2 \times \frac{4}{5}\] \[T_{3} = T_{2} \times \frac{4}{5}\] \[T_{4} = T_{3} \times \frac{4}{5}\] hope this helps

  3. rosedewittbukater
    • 3 years ago
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    @campbell_st can I ask you how you got 1/5 for r? Because don't you need to divide 10/2 and then subtract 1 to get 4, then switch the negatives to get -4?

  4. campbell_st
    • 3 years ago
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    ok the sum is 10 and the 1st term is 2 \[10 = \frac{2}{(1 - r)}.... so.... 10(1 - r) = 2......then ...... (1 -r) = \frac{2}{10}.... is the same as.... 1 - r = \frac{1}{5}\]

  5. rosedewittbukater
    • 3 years ago
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    Oh ok thanks. I always thought that you could just divide both sides by 2 but I just realized that it would end up as 5 = 1/(1-r). Thanks for your help!!

  6. rosedewittbukater
    • 3 years ago
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    One more question... How would it eventually add up to 10? I added up the first 4 terms and got about 5.9. And if it's infinite, how do we know the sum if it goes on forever? Sorry, this whole unit of sequences and series has been confusing me.

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