the arithmetic mean of the nine numbers in the given set (9,99,999......999999999) is a 9 digit number N all whose digits are distinct .The number N does not contain the digit
A) 0
B) 2
C) 5
D) 9

Hi  Amulya,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Arithmetic} \mathrm{mean}\left(\mathrm{A}.\mathrm{M}.\right)=\frac{\mathrm{Sum} \mathrm{of} \mathrm{given} \mathrm{quantities} }{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{quantities}} \\ \Rightarrow \mathrm{For} \mathrm{given} \mathrm{question} \\ \mathrm{A}.\mathrm{M}.=\frac{9+99+999+........+999999999}{9} \\ =9.\left(\frac{1+11+111+........+111111111}{9}\right) \\ =1+11+111+.....+111111111 \\ \mathrm{Now} 1+11=\mathbf{12} \left(\mathrm{Sum} \mathrm{of} \mathrm{first} \mathbf{two} \mathrm{terms}\right) \\ 1+11+111=\mathbf{123}\mathbf{ } \left(\mathrm{Sum} \mathrm{of} \mathrm{first} \mathbf{three} \mathrm{terms}\right) \\ 1+11+111+1111=\mathbf{1234} \left(\mathrm{Sum} \mathrm{of} \mathrm{first} \mathbf{four} \mathrm{terms}\right) \\ \mathrm{Similarly} \mathrm{sum} \mathrm{of} \mathbf{nine} \mathrm{terms} =\mathbf{123456789} \\ \Rightarrow \mathrm{A}.\mathrm{M}.=123456789 \\ \mathrm{Hence} \mathrm{A}.\mathrm{M}. \mathrm{does} \mathrm{not} \mathrm{contain} 0 \mathrm{as} \mathrm{one} \mathrm{of} \mathrm{its} \mathrm{digit}. \\ \mathbf{Hence}\mathbf{ }\mathbf{option}\left(\mathbf{A}\right)\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{correct}\mathbf{.} \end{gathered}$$

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