The average of n numbers is a. The first numb | vedclass

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(D) Let the $n$ numbers be $x_1, x_2, \dots, x_n$. Given that their average is $a$,so $\frac{\sum x_i}{n} = a$,which implies $\sum x_i = na$.The numbe

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$a + 2 \frac{2^n - 1}{n}$

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(D) Let the $n$ numbers be $x_1, x_2, \dots, x_n$. Given that their average is $a$,so $\frac{\sum x_i}{n} = a$,which implies $\sum x_i = na$.The numbers are increased by $2, 4, 8, \dots, 2^n$. This sequence is a Geometric Progression ($G$.$P$.) where the first term $A = 2$ and the common ratio $r = 2$.The sum of these increases is $S_n = \frac{A(r^n - 1)}{r - 1} = \frac{2(2^n - 1)}{2 - 1} = 2(2^n - 1)$.The sum of the new numbers is $\sum x_i + S_n = na + 2(2^n - 1)$.The new average is $\frac{na + 2(2^n - 1)}{n} = a + \frac{2(2^n - 1)}{n}$.

The average of $n$ numbers is $a$. The first number is increased by $2$,the second one is increased by $4$,the third one is increased by $8$,and so on. The average of the new numbers is:

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