Q162 / 270 · NDA 2011 (Phase 1) · Mathematics

Which term of the series $\frac{1}{4},-\frac{1}{2}, 1, \ldots$ is -128 ?

  • A
    9th
  • Correct answer
    B
    10 th
  • C
    11 th
  • D
    12 th

Solution

Since that given series are in G.P Where first term $=\frac{1}{4}$, common ratio $=-2$ Let -128 is $n$th term $\begin{array}{rlrl} & \therefore & -128 & =\frac{1}{4} \cdot(-2)^{n-1} \\ & & -128 \times 4 & =(-2)^{n-1} \\ \Rightarrow & & (-2)^9 & =(-2)^{n-1} \\ \Rightarrow & n & =9+1=10^{\text {th }} \end{array}$

Hence, the correct answer is option: 'B'.

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