Set K contains every multiple of 6 from 18 to 306 inclusive. If w is the median of set K, and x is the average (arithmetic mean) of set K, what is the value of w-x?	Choose one
A) -6
B) -3
C) 0
D) 3
E) 6
Set K ={18,24,30,....306}   Let the no. of terms be n.   so we know the AP formulae to calculate the the tn term:   tn= a+(n-1)*d where d = common difference of the terms.   306= 18 +(n-1)*6   n=49.   So the set K consist of n terms.   The median of the set having odd nos of elements is (n+1)/2., which in this case is 25.   Let's find the 25th term using the same formulae again:   t25= 18+(25-1)*6   t25= 162   So , the median of the set K is 162 i.e. w=162.   Now. lets find the average (arithmetic mean) of the set. For that we need to find the sum of all the elements first, lets call it S.   Since, set k is nothing but a Arithmetic Progression series having first element(a) as 18, common difference (d) as 6 and no. of terms(n) as 49.   Using the formulae to calculate sum of an AP series, which is   S= n/2[2a + (n-1)*d], we will calculate the sum.   so, S= 49/2 [2*18 + (49-1)*6]   This gives us S= 7938.   Now Arithmetic mean of Set K = 7938/no. of terms= 7938/49= 
<gadget id="calculator">7938/49</gadget>
<output>162</output>
162.   So x= 162.   Now, (w-x) = (162-162)= 
<gadget id="calculator">(162-162)</gadget>
<output>0</output>
0.   Therefore 0orrect Answer= 0
<result>C</result>
Set K contains every multiple of 6 from 18 to 306 inclusive. If w is the median of set K, and x is the average (arithmetic mean) of set K, what is the value of w-x?