Neildvesterdidiidn
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In a geometric sequence, the seventh term exceeds the fifth term by 1,920. Find the sum of
the first eleven terms if the common ratio of the sequence is 2. (81880)​

Sagot :

Jenaaaaaaaaaaaidn
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


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