Question

Let $nϵN$ & the A.M., G.M., H.M. & the root mean square of $n$ numbers $2n+1,2n+2,...,$ up to $n^{th}$ number are $A_n$, $G_n$, $H_n$ and $R_n$ respectively.
On the basis of above information answer the following questions$\underset{n\to \infty }{lim}\log \left(\frac{G_n}{n}\right)$ equals

• A. $\log \left(\frac{27}{4}\right)$
• B. $\log \left(\frac{27}{4e}\right)$
• C. $\log \left(\frac{27e}{4}\right)$
• D. $\log \left(\frac{4e}{27}\right)$

Answer 1

The correct option is B $\log \left(\frac{27}{4e}\right)$

$G_n=$ G.M. of $2n+1,2n+2,...,2n+n$ $={\{(2n+1)(2n+2)...(2n+n)\}}^{1/n}$
$\therefore$ $\frac{G_n}{n}=\frac{1}{n}{\{(2n+1)(2n+2)...(2n+n)\}}^{\frac{1}{n}}$

$\Rightarrow \frac{G_n}{n}={\left\{\frac{(2n+1)(2n+2)...(2n+n)}{n^n}\right\}}^{\frac{1}{n}}$
$\therefore$ $\log \left(\frac{G_n}{n}\right)=\frac{1}{n}\times \{\log (2n+1)+\log (2n+2)+...+\log (2n+n)...-\log n^n\}$

$\Rightarrow \log \left(\frac{G_n}{n}\right)=\frac{1}{n}\{\log (2n+1)+\log (2n+2)+...+\log (2n+n)...-n\log n\}$

$\Rightarrow \log \left(\frac{G_n}{n}\right)=\frac{1}{n}\{(\log (2n+1)-\log n)+(\log (2n+2)-\log n)+...+(\log (2n+n)-\log n)\}$

$\Rightarrow \log \left(\frac{G_n}{n}\right)=\frac{1}{n}\{\log (2+\frac{1}{n})+\log (2+\frac{2}{n})+...+\log (2+\frac{n}{n})\}$

$\Rightarrow \log \left(\frac{G_n}{n}\right)=\frac{1}{n}\sum _{r=1}^n\log (2+\frac{r}{n})$
$\therefore$ $\underset{n\to \infty }{lim}\log \left(\frac{G_n}{n}\right)=\underset{n\to \infty }{lim}\sum \frac{1}{n}\log (2+\frac{r}{n})$

${\int }_0^1\log (2+x)dx={((x+2)\log (2+x)-x)}_0^1$

$=(3\log 3-1)-2\log 2$

$=3\log 3-2\log 2-1$

$=\log \left(\frac{27}{4e}\right)$
Ans: B