Question

The arithmetic mean of $n$ numbers is $x$. If the sum of the first $(n-1)$ numbers is $k$, then the $n$th number is?

• A. $n\overline{x}+k$
• B. $n\overline{x}-k$
• C. $x-\frac{k}{n}$
• D. $\overline{x}+\frac{k}{n}$

Answer 1

Answer: (B)

$n\overline{x}-k$

Given that mean of $n$ numbers is $\overline{x}$

Let the numbers be $a_1,a_2,a_3,...,a_{n-1},a_n$

Therefore the mean $=\frac{a_1+a_2+a_3+...+a_{n-1}+a_n}{n}=\overline{x}$

$⟹a_1+a_2+a_3+...+a_{n-1}+a_n=n\overline{x}$ ————(1)

Given that sum of $(n-1)$ numbers is $k$

Therefore the sum $=a_1+a_2+a_3+...+a_{n-1}=k$ ————(2)

Substituting (2) in (1) we get

$k+a_n=n\overline{x}⟹a_n=n\overline{x}-k$

Therefore the $n^{th}$ number is $n\overline{x}-k$