A) \[p\]
B) \[3p\]
C) \[p+1\]
D) \[3p+1\]
Correct Answer: D
Solution :
Given that, \[p=\frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{n}}}{n}\] or \[pn={{x}_{1}}+{{x}_{2}}+{{x}_{3}}+.....+{{x}_{n}}\] \[\Rightarrow \] \[3pn=3{{x}_{1}}+3{{x}_{2}}+3{{x}_{3}}+......+3{{x}_{n}}\] \[\Rightarrow \]\[3p=\frac{3{{x}_{1}}+3{{x}_{2}}+3{{x}_{3}}+.....+3{{x}_{n}}}{n}\] ....(i) Now, \[\frac{3{{x}_{1}}+1+3{{x}_{2}}+1+3{{x}_{3}}+1+....+3{{x}_{n}}+1}{n}\] \[=\frac{3{{x}_{1}}+3{{x}_{2}}+3{{x}_{3}}+....+3{{x}_{n}}+n}{n}\] \[=\frac{3{{x}_{1}}+3{{x}_{2}}+3{{x}_{3}}+....+3{{x}_{n}}}{n}+1\] \[=3p+1\] [From (i)]
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