A) \[\frac{n+1}{2}\]
B) \[\frac{2n+1}{2}\]
C) \[\frac{2n+1}{3}\]
D) \[\frac{(2n+1)(n+1)}{6}\]
Correct Answer: A
Solution :
Let the weight of each number be\[w\]. Then, weighted mean, \[\bar{x}=\frac{1\times w+2\times w+3\times w+...+n\times w}{w+w+w+...+w}\] \[=\frac{w(1+2+3+...+n)}{w(1+1+1+...+1)}\] \[=\frac{w\frac{n(n+1)}{2}}{wn}=\frac{n+1}{2}\]
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