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anonymous

  • one year ago

Find the sum of a 9-term geometric sequence when the first term is 4 and the last term is 1,024.

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  1. anonymous
    • one year ago
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    @mathmate

  2. anonymous
    • one year ago
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    @Preetha

  3. mathmate
    • one year ago
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    Here's an article to familiarize yourself with geometric sequences and sums. If you encounter problems while solving, please post again. Hint: in a geometric sequence, \(ar^0=a\) is the first term, r is the common ratio, and \(ar^{n-1}\) is the nth term. So the nineth term divided by the first term equals \(\Large \frac{ar^{9-1}}{ar^{0}}=r^8\) This will allow you to solve for r.

  4. anonymous
    • one year ago
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    why thank you, let me try it out and I'll ask you for help if needed c:

  5. mathmate
    • one year ago
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    Here's the link that I forgot to post: http://www.mathsisfun.com/algebra/sequences-sums-geometric.html

  6. anonymous
    • one year ago
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    alright so I have a 113.77778 added to the last term but it comes out to 1027 in the ninth term. Could I get some guidance here?

  7. mathmate
    • one year ago
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    If you solve r^8=1024/4, you should find an integer value for r (whole number). The rest will work well from there.

  8. anonymous
    • one year ago
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    that means r = 2, but that makes the 8th term 1,024

  9. anonymous
    • one year ago
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    things got all out of wack, sorry, I should have tried counting better, it was r = 2 the whole time. Thank you for your time.

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