Question

If $x_1,x_2,x_3,....$ are n values of the variable x, then mathematical averages, Arithmetic Mean (A.M.), Geometric Mean (G.M) and Harmonic Mean (H.M) are count by the following formulas's.

$A.M.=\frac{x_1+x_2+x_3+...+x_n}{n}=\frac{1}{n}\left\{\sum _{i=1}^n{x}_1\right\}\nless x_1>0$

$G.M.={(x_1\cdot x_2...x_n)}^{1/n}$ if each $x_1(i=1,2,....,n)$ is positive and

$H.M=\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}}=\frac{1}{\frac{1}{n}{\sum }_{i=1}^n\left(\frac{1}{x_1}\right)}$

* In case of frequency distribution $x/f_1(i=1,2,...,n)$ where $f_1$ is the frequency of the variable $x_r$ then the calculation of A.M.is counted as $A.M=\sum _{i=1}^n{f}_1{x}_1/NwhereN=f_1+f_2....+f_n$

** If $w_1,w_2,....,w_{{}_n}$ be the weight assigned to the n values

$x_1,x_2,....,x_{{}_n}$ then weighted A.M is counted by $\frac{{\sum }_{i=1}^n{w}_1{x}_1}{{\sum }_{i=1}^n{w}_1}$

*** if $G_1,G_2$ are the G. M's of two series of sizes $n_1$ & $n_2$ respectively, then the geometric mean (G M.) of the combined series is counted by $\log (G.M)=\frac{n_1\log G_1+n_2\log G_2}{n_1+n_2}$

On the basis of above information answer the following questions,Consider the series 1,4,16,64,256, ......,$4^n$, then which of the following is not true?

• A. $A.M.=\frac{4^{n+1}-1}{3(n+1)}$
• B. $G.M.=2^n$
• C. $H.M.=\frac{3\cdot 4^n(n+1)}{4^{n+1}-1}$
• D. A.M.$=$G.M.$=$H.M.

Answer 1

The correct option is D A.M.$=$G.M.$=$H.M.

Given numbers are $4^0$, $4^1$, $4^2$, ..., $4^n$. So there are (n+1) numbers
$\therefore$ $A.M.=\frac{4^0+4^1+4^2+...+4^n}{n+1}$
$=\frac{4^{n+1}-1}{(n+1)(4-1)}$ {$S_n$ of $G.P.=\frac{a(r^n-1)}{r-1}$}
$=\frac{4^{n+1}-1}{3(n+1)}$

G.M. of $x_1$, $x_2$, ..., $x_n$ is ${(x_1{x}_2...x_n)}^{\frac{1}{n}}$
$\therefore$ G.M. of $(4^0,4^1,4^2,...,4^n)={(4^0{.4}^1{.4}^2....{.4}^n)}^{\frac{1}{n+1}}$
$=\left(4^{\frac{n(n+1)}{2}}\right)\frac{1}{n+1}$
$=4^{\frac{n}{2}}={\left(4^{1/2}\right)}^n$
$=2^n$

Again H.M. for (n+1) numbers $x_1$, $x_2$, ..., $x_{n+1}$ is given by
$H.M.=\frac{n+1}{\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_{n+11}}}$

$\therefore$ H.M. of $1,4,4^2,...,4^n$ is
$=\frac{n+1}{1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^n}}$
$=\frac{4^n(n+1)}{4^n+4^{n-1}+...+1}$
$=\frac{{3.4}^n(n+1)}{4^{n+1}-1}$

now A.M., G.M., H.M. are all different

$\therefore$ Choice (d) is the final choice, which is not true.