Geomatic Sequence
posted by Nina
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.

Steve
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 
Reiny
ar^6  ar^4=1920
64a16a=1920
48a=1920
a = 40
sum(11) = 40(2^11  1)/(21) = 81880 
Steve
oops  good catch, Reiny. You actually answered the question :)

stephanie
find the fifth term in the sequence a1=4a2=8a3=16 can you guys help me
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