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 $a+\frac{2(2^{n}–1)}{n}$.