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

NDA 2011 · Mathematics · Sequences and Series

If n!, 3 ×(n!) and (n+1)! Are in GP, then the value of n will be

If n!,3×(n!)n!, 3 \times(n!) and (n+1)!(n+1)! Are in GP, then the value of nn will be

Options

  1. A33
  2. B44
  3. C88
  4. D1010

Correct answer

C. 88

Step-by-step solution

n!,3(n!)and(n+1)!n!, 3(n!) and (n+1)! are in G.P. ∴[3(n!)]2=n!×(n+1)!\therefore \quad[3(n!)]^2=n!\times(n+1)!9(n!)(n!)=n!(n+1)⋅n!9=n+1\begin{aligned} 9(n!)(n!) & =n!(n+1) \cdot n! \\ 9 & =n+1\end{aligned}⇒n=8\Rightarrow \quad n=8.
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