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NDA 2011
Exam
NDA
Year
2011
Subject
Mathematics
Chapter
Sequences and Series

NDA 2011 · Mathematics · Sequences and Series

Which term of the series (1)/(4),-(1)/(2), 1, is -128 ?

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

Options

  1. A9th
  2. B10 th
  3. C11 th
  4. D12 th

Correct answer

B. 10 th

Step-by-step solution

Since that given series are in G.P Where first term =14=\frac{1}{4}, common ratio =−2=-2 Let -128 is nnth term ∴−128=14⋅(−2)n−1−128×4=(−2)n−1⇒(−2)9=(−2)n−1⇒n=9+1=10th \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}
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