If p and n are positive integers, with p > n, what is the average of n consecutive multiples of p (starting with p), less the average of p consecutive multiples of n (starting with n)?  Choose one.
A) (p + n)/2 B) (p – n)/2 C) p + n D) p – n E) p^2 – n^2
Let, n = 2   and p = 3   i.e. Average of 2 (n) consecutive multiples of 3 (p) starting from 3 (p) = (3+6)/2 = 9/2   and Average of 3 (p) consecutive multiples of 2 (n) starting from 2 (n) = (2+4+6)/3 = 12/3= 
<gadget id="calculator">12/3</gadget>
<output>4</output>
4   The average of n consecutive multiples of p (starting with p), less the average of p consecutive multiples of n (starting with n)= (9/2)-4 = (1/2)   A. (p + n)/2 = (3+2)/2 = 7/2 INCORRECT   B. (p – n)/2 = (3-2)/2 = 1/2 CORRECT   C. p + n = (3+2) = 7 INCORRECT   D. p – n = (3-2)= 
<gadget id="calculator">(3-2)</gadget>
<output>1</output>
1 INCORRECT   E. p^2 – n^2= 3^2 - 2^2 = 5 INCORRECT   Answer: option (p – n)/2
<result>B</result>
If p and n are positive integers, with p > n, what is the average of n consecutive multiples of p (starting with p), less the average of p consecutive multiples of n (starting with n)?