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Mathematics

The 11th term of the G.P. 18,14,2,1,....\dfrac{1}{8}, -\dfrac{1}{4}, 2, -1, …. is

  1. 64

  2. -64

  3. 128

  4. -128

AP GP

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Answer

Given, a = 18\dfrac{1}{8} and common ratio = r = 1418=2.\dfrac{-\dfrac{1}{4}}{\dfrac{1}{8}} = -2.

We know that

    an = arn - 1

a11=18(2)111=18(2)10=18×210=21023=27=128.\therefore a_{11} = \dfrac{1}{8}(-2)^{11 - 1} \\[1em] = \dfrac{1}{8}(-2)^{10} \\[1em] = \dfrac{1}{8} \times 2^{10} \\[1em] = \dfrac{2^{10}}{2^3} \\[1em] = 2^7 \\[1em] = 128.

Hence, Option 3 is the correct option.

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