Asked 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.
Answered by 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
Answered by Reiny
ar^6 - ar^4=1920
64a-16a=1920
48a=1920
a = 40

sum(11) = 40(2^11 - 1)/(2-1) = 81880
Answered by Steve
oops - good catch, Reiny. You actually answered the question :-)
Answered by stephanie
find the fifth term in the sequence a1=-4a2=8a3=-16 can you guys help me
Answered by Rowena
find the first term if the 6th and 7th term are 320 and 1920 respectively.

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