Ratios & Proportion
Aptitude

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Q.

The ratio of squares of first $n$ natural numbers to square of sum of first $n$ natural numbers is $17:325$. The value of $n$ is:

 A.

15

 B.

25

 C.

35

 D.

40

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Solution:
Option(B) is correct

Sum of squares of first $n$ natural numbers

$=\sum{n^2}=$ \(\dfrac{n(n+1)(2n+1)}{6}\)

Squares of sum of first $n$ natural numbers 

$= (\sum{n})^2=$\(\dfrac{n(n+1)}{2}\times \dfrac{n(n+1)}{2}\)

Now the ratio is

\(\dfrac{n(n+1)(2n+1)}{6}\) : \(\dfrac{n(n+1)}{2}\times \dfrac{n(n+1)}{2}\)$=17:325$

$\Rightarrow \dfrac{\left(\frac{2n+1}{3}\right)}{\left(\frac{n(n+1)}{2}\right)}=\dfrac{17}{325}$

$\Rightarrow \dfrac{2(2n+1)}{3n(n+1)}=\dfrac{17}{325}$

$\Rightarrow 650(2n+1)=51n(n+1)$

$\Rightarrow 1300n+650=51n^2+51n$

$\Rightarrow 51n^2-1300n+51n-650=0$

$\Rightarrow 51n^2-1249n-650=0$

Upon solving the above equation, we get,

$n=25, -0.5098$

Out of these two, $n=\textbf{25}$ is there in the solution. 

Thus option (B) is the right choice.


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