Average of n numbers is a. The first number is increased by 2, second one is increased by 4, the third one is increased by 8 and so on. The average of the new numbers is
Options:
A . $a+(2^(n-1)-1)/n$
B . $a+2(2^n-1)/n$
C . $a+2^(n-1)/n$
D . $a+(2^n-1)/n$
Answer: Option B Answer: (b) Sum of new numbers = na + (2 + 4 + 8 + 16 ..... to n terms) Now, S = 2 + 4 + 8 + 16 + ..... to n terms Here, a = first term = 2 r = common ratio = $4/2$ = 2 It is a geometric series. ∴ S =${a({r^n} - 1)}/{r-1}$= ${2({2^n} - 1)}/{2-1}$ = 2 ($2^n$ –1) ∴ Required average = ${na +2({2^n}-1)}/n$ a +${2({2^n}-1)}/n$
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