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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Sequences and Series

NDA 2022 · Mathematics · Sequences and Series

Let x be the mean of squares of first n natural numbers and y be...

Let xx be the mean of squares of first nn natural numbers and yy be the square of mean of first nn natural numbers. If xy=5542\frac{x}{y}=\frac{55}{42}, then what is the value of nn ?

Options

  1. A2424
  2. B2525
  3. C2727
  4. D3030

Correct answer

C. 2727

Step-by-step solution

Mean of squares of first nn natural number x=n(n+1)(2n+1)6n=(n+1)(2n+1)6x=\frac{n(n+1)(2 n+1)}{6 n}=\frac{(n+1)(2 n+1)}{6} Square of mean of first nn natural numbers y=(n(n+1)2n)2=(n+1)24∴xy=2(2n+1)3(n+1)=5542⇒28(2n+1)=55(n+1)⇒n=27\begin{aligned} & y=\left(\frac{n(n+1)}{2 n}\right)^2=\frac{(n+1)^2}{4} \\ & \therefore \frac{x}{y}=\frac{2(2 n+1)}{3(n+1)}=\frac{55}{42} \\ & \Rightarrow 28(2 n+1)=55(n+1) \Rightarrow n=27 \end{aligned}
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