Geomatic Sequence

In a geomatic sequence, the seventh term exceeds the fifth term by 1920. Find the sum of the first eleven terms if the common ratio of the sequence is 2.

asked by Nina
  1. that's geometric sequence

    T7 = T5*r^2 = T5+1920
    r = 2, so

    4*T5 = T5 + 1920
    3T5 = 1920
    T5 = 640

    640 = 2^4 * 40

    So, a=40 and the sequence is

    40 80 160 320 640 1280 2560 5120 10240 20480 40960 ...

    Note that 2560 = 640 + 1920

    posted by Steve
  2. ar^6 - ar^4=1920
    64a-16a=1920
    48a=1920
    a = 40

    sum(11) = 40(2^11 - 1)/(2-1) = 81880

    posted by Reiny
  3. oops - good catch, Reiny. You actually answered the question :-)

    posted by Steve
  4. find the fifth term in the sequence a1=-4a2=8a3=-16 can you guys help me

    posted by stephanie

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