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Find the 10th term of the following geometric sequence: 2, 8, 32, 128.     
A. 248,221
B. 621,325
C. 164,357
D. 524,288

Respuesta :

Earmijo1
the 10th term of the following geometric sequence is D. 524,288
its a geometric sequence, the common ratio = 8/2 = 32/8 = 128/32 = 4

so the nth term = (first term)*CR^(n-1)

= 2*4^(n-1)

substituting n=10

2*4^(10-1)

= 2*4^9

= 524,288

ans is D

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