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1 June, 10:35

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

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Answers (2)
  1. 1 June, 12:59
    0
    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
  2. 1 June, 13:06
    0
    The 10th term of the following geometric sequence is D. 524,288
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