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 –
A. $\Large a+\frac{2\left(2^{n}-1\right)}{n}$ B. $\Large a+\frac{2^{n+1}-1}{n}$ C. $\Large a+\frac{2^{n+1}}{n}$ D. $\Large a+2 \frac{2^{n-1}}{n}$ Answer: Option AShow Answer
Solution(By Apex Team)
Series:- a, a + 2, a + 4…..
Sum = na + 2 + 4 + ….. upto n terms
Sum = na + Sn
$\Large S_{n}=\frac{2\left(2^{n}-1\right)}{2-1}$
Average
$\Large S_{n}=a+\frac{2\left(2^{n}-1\right)}{2-1}$
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