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(A) $34$ (B) $42$ (C) $38$ (D) $40$

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1) First one is definition of mean i.e. mean is equal to sum of observations divided by number of observations.

2) And the second one is Arithmetic progression. It is a sequence of numbers such that the difference between the consecutive terms is constant.

The mean of the given set ${x_1},{x_2},...{x_{10}}$ is$20$.

We know that the formula for mean $m = \dfrac{{{x_1} + {x_2} + {x_3}......... + {x_n}}}{n}$ where, ${x_1},{x_2},...{x_n}$ are the observations and $n$ is number of observations. By applying the formula,

\[\dfrac{{({x_1} + {x_2} + ... + {x_{10}})}}{{10}} = 20\] [As given that mean is $20$]

\[ \Rightarrow {x_1} + {x_2} + ... + {x_{10}} = 200\] equation (1)

We have to find the mean of ${x_1} + 4,{x_2} + 8,{x_3} + 12,...{x_{10}} + 40$, therefore,

$\dfrac{{({x_1} + 4 + {x_2} + 8 + {x_3} + 12 + ... + {x_{10}} + 40)}}{{10}}$

$ = \dfrac{{({x_1} + {x_2} + {x_3} + ... + {x_{10}}) + (4 + 8 + 12 + ... + 40)}}{{10}}$

\[ = \dfrac{{200 + (4 + 8 + 12 + ... + 40)}}{{10}}\] [from equation (1)] equation (2)

We can observe that $4 + 8 + 12 + ... + 40$ is an Arithmetic progression,

$8 - 4 = 4$

$12 - 8 = 4$

It means that the difference between two consecutive terms of the sequence is common i.e. $4$.

We know that, formula for sum of finite number of terms of an A.P. is

${S_n} = \dfrac{n}{2}\left\{ {2a + (n - 1)d} \right\}$

Where $a$ is first term, $d$ is common difference and $n$ is total number of term in the A.P.

For $4 + 8 + 12 + ... + 40$,

$a = 4$, $d = 4$ and $n = 10$

On substituting the values,

${S_{10}} = \dfrac{{10}}{2}\left\{ {2 \times 4 + \left( {10 - 1} \right)4} \right\}$

$ \Rightarrow {S_{10}} = 5\left\{ {8 + 9 \times 4} \right\}$

$ \Rightarrow {S_{10}} = 5\left\{ {8 + 36} \right\}$

$ \Rightarrow {S_{10}} = 5\left\{ {44} \right\}$

$ \Rightarrow {S_{10}} = 220$

$\therefore 4 + 8 + 12 + ... + 40 = 220$

On substituting the value in equation (2),

\[\dfrac{{({x_1} + 4 + {x_2} + 8 + {x_3} + 12 + ... + {x_{10}} + 40)}}{{10}} = \dfrac{{200 + 220}}{{10}}\]

$ \Rightarrow \dfrac{{({x_1} + 4 + {x_2} + 8 + {x_3} + 12 + ... + {x_{10}} + 40)}}{{10}} = \dfrac{{420}}{{10}}$

$ \Rightarrow \dfrac{{({x_1} + 4 + {x_2} + 8 + {x_3} + 12 + ... + {x_{10}} + 40)}}{{10}} = 42$

Hence the mean of ${x_1} + 4,{x_2} + 8,{x_3} + 12,...,{x_{10}} + 40 = 42$

Mean = $\overline x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}{f_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}$ where ${f_i}$ is the frequency of ${i_{th}}$ observation ${x_i}$.

Formula for the mean for grouped data when class intervals are given

Mean = $\overline x = \dfrac{{\sum\limits_{i = 1}^n {{x_i}{f_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}$ where ${f_i}$ is the frequency of ${i_{th}}$ class whose class mark is ${x_i}$.

Class mark $ = $ (Upper class limit $ + $ Lower class limit)$/2$