Here, $n=2$

For any given numbers, their means, Arithmetic mean (AM), Geometric mean, (GM) and Harmonic mean are related by the relation
$ GM^2 = AM \times HM $

$\text{Arithmetic mean for ungrouped data:-}$ Arithmetic mean for ungrouped data is the average of the values in the data set. i.e., sum of the values in the data set divided by the total number of values in the data set.

Consider the following table, to calculate mean:

Consider the following table, to calculate mean:

Answer:- Taking continuous class interval with width $= 10$

$ \therefore $ Mean $= 48.4$

Answer: Using direct method to calculate mean :

Consider the following table, to calculate mean:

Consider the following table, to calculate mean:

Consider the following table, to calculate mean:

Given:- $\Sigma f = 52 $

Also $ \Sigma f _i = 33  f _1 + f _2 = 52 $

Formula of finding Mean using step deviation method is

$\displaystyle\text{Let } x _1,x _2,x _3,x _4\text{ be four observations.}$
$\displaystyle\text{According to question}$
$\displaystyle \frac{ x _1+x _2+x _3+x _4}{4}=20$
$\Rightarrow \displaystyle { x _1+x _2+x _3+x _4}=80$
$\displaystyle\text{After adding 'c' to each observation the new A.M becomes 22.}$
$\Rightarrow \displaystyle \frac{ (x _1+c)+(x _2+c)+(x _3+c)+(x _4+c)}{4}=22$
$\Rightarrow \displaystyle  (x _1+x _2+x _3+x _4)+4c=88$
$\Rightarrow \displaystyle  80+4c=88$
$\Rightarrow \displaystyle  4c=8$
$\Rightarrow \displaystyle  c=2$
Options B is correct.

Harmonic mean of two numbers $ a $ and $ b = \dfrac {2ab}{a+b} $

So, harmonic mean of $ 3\ and\  5 $ is $ \dfrac {2 \times 3 \times 5}{3+5} = \dfrac {15}{4} $

Geometric mean or GM of two numbers $ a $ and $ b $ is $ \sqrt {ab} $
So, GM of $ 4\  and\  64 $ is $ \sqrt {4 \times 64} = 2 \times 8 = 16 $

Harmonic mean of two numbers $ a $ and $ b = \dfrac {2ab}{a+b} $

So, harmonic mean of $ 20\  and\  30 $ is $ \dfrac {2 \times 20 \times 30}{20+30} = 24 $

Given that, there are $5$geometric means between the two numbers $\dfrac{1}{3}$and $243$ , we have to find $7=\left( 5+2 \right)$

terms in G.P. of which $\dfrac{1}{3}$ is the first, and$243$ the seventh. Let r be the common ratio;

then $243$  = the seventh term =$\left( \dfrac{1}{3} \right){{r}^{\left( 7-1 \right)}}=\dfrac{1}{3}.{{r}^{6}}$.

Therefore,${{r}^{6}}=3.x.243={{3.3.3}^{4}}={{3}^{6}}$;

whence $r=6$

and the series is$\dfrac{1}{3},1,3,9,27,81,243$

(using the standard form a, ar, ar², ar³ …… of a G.P. ).

Now, the geometric mean between two given quantities$a,b=\sqrt{ab}$

Therefore, the required geometric means are,

$ \sqrt{\dfrac{1}{3}.x.3},\sqrt{1.x.9},\sqrt{3.x.27},\sqrt{9.x.82},\sqrt{27.x.243} $$

$ =1,3,9,27,81 $$

Therefore, the sum of the $5$  geometric means is

$1+3+9+27+81=121$

Hence, this is the answer.

Geometric mean of $n$ numbers $=(\prod _{i=1}^{n} x _{i})^{1/n}$

$\begin{array}{l} \left( { 5+\sqrt { 2 }  } \right) { x^{ 2 } }-\left( { 4+\sqrt { 5 }  } \right) x+8+2\sqrt { 5 } =0 \ a=5+\sqrt { 2 }  \ b=-\left( { 4+\sqrt { 5 }  } \right)  \ c=8+2\sqrt { 5 }  \ Harmonic\, \, mean\, \, of\, \, \lambda ,\beta  \ =\frac { { 2\lambda \beta  } }{ { \lambda +\beta  } }  \ \lambda \beta =\frac { c }{ a } =\frac { { 8+2\sqrt { 5 }  } }{ { 5+\sqrt { 2 }  } }  \ \lambda +\beta =\frac { { -b } }{ a } =\frac { { 4+\sqrt { 5 }  } }{ { 5+\sqrt { 2 }  } }  \ Harmonic\, \, mean=\,  \ \frac { { 2\frac { { \left( { 8+2\sqrt { 5 }  } \right)  } }{ { 5\sqrt { 2 }  } }  } }{ { \frac { { 4+\sqrt { 5 }  } }{ { 5+\sqrt { 2 }  } }  } }  \ =\frac { { 2\left( { 8+2\sqrt { 5 }  } \right)  } }{ { 4+\sqrt { 5 }  } }  \ =4\, \, \,  \end{array}$

Given $ 30 + {f} _{1} +{f} _{2} = 50 $
$ => {f} _{1} +{f} _{2} = 20 $   -- (1)

Given, Mean $ = \cfrac { \sum { fx }  }{ \sum { f }} =62.8 $
$ => \cfrac { 2060  + 30{f} _{1}+70{f} _{2}}{30 + {f} _{1} +{f} _{2}} = 62.8 $

$ =>  2060  + 30{f} _{1}+70{f} _{2} = 1884 + 62.8{f} _{1} + 62.8{f} _{2} $ 

$ 32.8{f} _{1} - 7.2{f} _{2} =176 $

=> $ 8.2{f} _{1} - 1.8{f} _{2} = 44 $

=> $ 4.1{f} _{1} - 0.9{f} _{2} = 22 $ -- (2)