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anonymous

  • one year ago

HEEEEEEEEEEELPPPP! The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to represent the first three terms, respectively.] The three numbers are _____, _____, and _____.

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  1. anonymous
    • one year ago
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    \[\frac{ a }{ r }*a*ar=-\frac{ 1 }{ 125 },a^3=\frac{ -1 }{ 125 }=\left( \frac{ -1 }{ 5 } \right)^3\] \[a=\frac{ -1 }{ 5 }\] \[terms~ are=\frac{ -1 }{ 5r },\frac{ -1 }{ 5 },\frac{ -r }{ 5 }\] \[-\frac{ 1 }{ 5r }-\frac{ 1 }{ 5 }-\frac{ r }{ 5 }=-\frac{ 7 }{ 10 }\] multiply each term by -10r and solve the quadratic formed for r complete it.

  2. anonymous
    • one year ago
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    So I would literally multiply -1/5r by -10r? If so, does that cancel out the r that is in the denominator? This was from a previous unit in my math class that just got re-assigned so I don't remember any of it.

  3. anonymous
    • one year ago
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    Or do I convert the denominators of the terms into -10r through finding the common denominator?

  4. anonymous
    • one year ago
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    @surjithayer?

  5. anonymous
    • one year ago
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  6. anonymous
    • one year ago
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    the idea is, it is free of denominator

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