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Given that 6,32,36,92,......

Given that 6,32,36,92,... are the first four terms of an exponential sequence (G.P), find in its simplest form the 8th term. 

A.

272

B.

276

C.

812

D.

816

Correct answer is C

Tn=arn1 (Geometric progression)

a=6,r=T2T1=326

r=186=3

= (\sqrt{6})(27\sqrt{3}) = 27\sqrt{18} = 81\sqrt{2}