Question

If ${\sum }_{n=1}^{n}n$, $\frac{\sqrt{10}}{3}{\sum }_{n=1}^n{n}^2$, ${\sum }_{n=1}^n{n}^3$ are in GP, then the value of $n$ is

• A. $2$
• B. $3$
• C. $4$
• D. nonexistent

Answer 1

The correct option is C $4$

Let the three terms in GP be $a,b,c$

Thus , $a={\sum }_{n=1}^{n}n=\frac{n(n+1)}{2}$

$b=\frac{\sqrt{10}}{3}{\sum }_{n=1}^n{n}^2=\frac{\sqrt{10}}{3}\frac{n(n+1)(2n+1)}{6}$

$c={\sum }_{n=1}^n{n}^3=(\frac{n(n+1)}{2}{)}^2$

Now, $b^2=ac$

$\Rightarrow \frac{10}{9}(\frac{n(n+1)(2n+1)}{6}{)}^2=(\frac{n(n+1)}{2}{)}^3$

$\Rightarrow \frac{10}{9}(\frac{2n+1}{6}{)}^2=\frac{1}{4}\frac{n(n+1)}{2}$

$\Rightarrow 20(4n^2+4n+1)=81(n^2+n)$

$\Rightarrow n^2+n-20=0$

$\Rightarrow (n+5)(n-4)=0\Rightarrow n=4$

$\therefore C$ is correct.