# 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.

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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