Question : 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:
Option 1: $a+\frac{2^{n–1}–1}{n}$
Option 2: $a+\frac{2(2n–1)}{n}$
Option 3: $a+\frac{2^{n–1}}{n}$
Option 4: $a+\frac{2^{n}–1}{n}$
Correct Answer: $a+\frac{2(2n–1)}{n}$
Solution :
Given: The average of $n$ numbers is $a$.
The sum of $n$ numbers is $n×a$.
According to the question,
Let S be the sum of increased value and $r$ = common ratio = $\frac{4}{2}= 2$
S = 2 + 4 + 8 + 16 .... upto $n$ terms.
It is a geometric series.
S = $\frac{a(r^{n}–1)}{r–1}=\frac{2(2^{n}–1)}{2–1}= 2(2^{n}–1)$
Now, the sum of old numbers and increased value is $(n×a)+2(2^{n}–1)$.
So, the required average = $\frac{na+2(2^{n}–1)}{n}=a+\frac{2(2^{n}–1)}{n}$
Hence, the correct answer is $a+\frac{2(2^{n}–1)}{n}$.
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