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Karyatou

  • 3 years ago

The first three terms of a sequence are given. 15,10,20/3,... Find the 6th term.

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  1. across
    • 3 years ago
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    Let \(a_0=15\). Then\[a_{n+1}=\frac{2a_n}{3}\qquad n=0,1,\dots\]

  2. across
    • 3 years ago
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    Here, you're looking for \(a_5\).

  3. across
    • 3 years ago
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    That is,\[a_5=(2/3)(2/3)(2/3)(2/3)(2/3)a_0=\frac{32}{243}a_0=\frac{480}{243}\approx1.975308641975309.\]

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