Problem:
By definition r!=r(r−1)⋯1 and (kj)=k!(j−k)!j!, where r,j,k are positive integers and k<j. If (1n),(2n),(3n) form an arithmetic progression with n>3, then n equals
Answer Choices:
A. 5
B. 7
C. 9
D. 11
E. 12
Solution:
Write
(1n)=n,(2n)=2n(n−1) and (3n)=6n(n−1)(n−2).
Hence
2n(n−1)−n=6n(n−1)(n−2)−2n(n−1)
Thus
n3−9n2+14n=0n(n−2)(n−7)=0
Since n>3,n=7 is the solution.
(The answer may also be obtained by evaluating the sequence (1n),(2n),(3n) for the values of n listed as choices.)