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    The sum of the first three terms of a geometric progression series of real numbers is 65 while the sum of the first six terms is 1820. Determine the tenth term of the series...

    I've tried to work this out but i am struggling, please help.

    Thanks
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    (Original post by belle-belle)
    The sum of the first three terms of a geometric progression series of real numbers is 65 while the sum of the first six terms is 1820. Determine the tenth term of the series...

    I've tried to work this out but i am struggling, please help.

    Thanks
    a + ar + ar^2 = 65
    65 + (ar^3 + ar^4 + ar^5) = 1820 => ar^3 + ar^4 + ar^5 = 1755

    a(1 + r + r^2) = 65 => a(r^3 + r^4 + r^5) = 65r^3
    => 65r^3 = 1755 => r^3 = 27 => r = 3

    a(1 + 3 + 9) = 65 => 13a = 65 => a = 5

    10th Term = ar^9 = 5(3^9) = 5(19683) = 98415
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    (Original post by Physics Enemy)
    a + ar + ar^2 = 65
    65 + (ar^3 + ar^4 + ar^5) = 1820 => ar^3 + ar^4 + ar^5 = 1755

    a(1 + r + r^2) = 65 => a(r^3 + r^4 + r^5) = 65r^3
    => 65r^3 = 1755 => r^3 = 27 => r = 3

    a(1 + 3 + 9) = 65 => 13a = 65 => a = 5

    10th Term = ar^9 = 5(3^9) = 5(19683) = 98415
    Thank you very much, i understand now.
 
 
 
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