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.

Answer 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

Answer 2

ar^6 - ar^4=1920
64a-16a=1920
48a=1920
a = 40

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

Answer 3

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

Answer 4

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

Answer 5

find the first term if the 6th and 7th term are 320 and 1920 respectively.