# Show that the Geometric Mean $g$ can be expressed as $\log g=a+\dfrac{k}{n}\sum_{i=1}^r f_i(i-1)$

In a frequency distribution the upper boundary of each class interval has a constant ratio to the lower boundary. Show that the Geometric Mean $g$ can be expressed as $$\log g=a+\dfrac{k}{n}\sum_{i=1}^r f_i(i-1)$$

where $a$ is the class mark of the first class, $k$ is the logarithm of the ratio between upper and lower boundary and $n=\sum_{i=1}^r i.$

My try:

$\displaystyle g=\left\{\left(\dfrac{l_i+u_i}{2}\right)^{f_i}\right\}^{\dfrac{1}{\displaystyle\sum f_i}}=\left\{\left(\dfrac{l_i+Rl_i}{2}\right)^{f_i}\right\}^{\dfrac{1}{\displaystyle\sum f_i}}=\left\{{l_i}^{f_i}\left(\dfrac{1+R}{2}\right)^{f_i}\right\}^{\dfrac{1}{\displaystyle\sum f_i}}$

Then $\log g={\dfrac{1}{\displaystyle\sum f_i}}\left[f_i\left(\log l_i+\log \left(\dfrac{1+R}{2}\right)\right)\right]$

I can't proceed further.

• I've added the tag self-study. Normally you are expected to show your ideas so far, not just ask us to do your homework for you. – Nick Cox Mar 9 '14 at 9:32
• Please see the tag wiki info. What have you tried? – Glen_b Mar 9 '14 at 9:40
• Is your expression for $n$ correct? – Glen_b Mar 9 '14 at 9:54
• Please explain what "$f_i$" means and tell us what $g$ is supposed to be the GM of. (I cannot find any interpretation of the "geometric mean" and of the $f_i$ that makes a formula like this generally true.) – whuber Mar 9 '14 at 23:34
• $f_i$ is the frequency of the $i^{th}$ class, $g$ is the gm of the class marks considering frequencies – s.jan Mar 10 '14 at 15:30

Quite a few errors in the entire post.

First of all, it would have made sense to say $n=\sum_{i=1}^r f_i$, the total frequency being the total number of observations for weighted means.

The derivation is plain algebraic manipulation to simply arrive at a formula.

Based on comments, $g$ is the weighted geometric mean of the class marks and $f_i$ is the frequency of the $i$th class (This should be added to the post itself).

Let $$\frac{\text{Upper class boundary}}{\text{Lower class boundary}}=c\qquad\text{for all classes}$$

$$x_i=\text{class mark of the ith class},\quad i=1,2,\cdots,r$$

$$f_i=\text{frequency of the ith class},\quad i=1,2,\cdots,r$$

Denote the $i$th class by $[a_i,a_{i+1}],\quad i=1,2,\cdots,r$.

We are given that $$\frac{a_2}{a_1}=\frac{a_3}{a_2}=\cdots=\frac{a_{r+1}}{a_r}=c$$

Now,

\begin{align} \frac{a_3}{a_2}&=\frac{a_2}{a_1} \\\implies \frac{a_3}{a_2}+1&=\frac{a_2}{a_1}+1 \\\implies \frac{(a_2+a_3)/2}{(a_1+a_2)/2}&=\frac{a_2}{a_1} \\\implies \frac{x_2}{x_1}&=\frac{a_2}{a_1} \end{align}

$$\frac{x_2}{x_1}=\frac{x_3}{x_2}=\cdots=\frac{x_{r+1}}{x_r}=c$$

From this we have $$x_i=x_1c^{\,i-1}\quad,i=1,2,\cdots,r$$

Thus,

\begin{align} g=\left(\prod_{i=1}^rx_i^{\,f_i}\right)^{1/n}&=\left\{\prod_{i=1}^r\left(x_1c^{\,i-1}\right)^{f_i}\right\}^{1/n} \\&=\left\{\prod_{i=1}^r x_1^{\,f_i}\,c^{\,(i-1)f_i}\right\}^{1/n} \end{align}

On simplification we get

\begin{align} \log g&=\frac{1}{n}\log\left[x_1^{\,\sum_{i=1}^rf_i}\,c^{\,\sum_{i=1}^r(i-1)f_i}\right] \\&=\frac{1}{n}\left[\log x_1^n+\log c^{\,\sum_{i=1}^r(i-1)f_i}\right] \\&=\frac{1}{n}\left[n\log x_1+\left\{\sum_{i=1}^r(i-1)f_i\right\}\log c\right] \\&=\log x_1+\frac{\log c}{n}\sum_{i=1}^r(i-1)f_i \end{align}

This is the expression we can arrive at. Thus in the formula asked to be proved in the original post, the $a$ should be the logarithm of the class mark of the first class, and not just the class mark.