challenging quest?

Question

if the sixth term in the expansionof (3/2+x/3)^n when x=3 is numerically greatest then the possible integral values of n can be

Answer 1

Note that when x = 3, x/3 = 1.

The sixth term is (n choose 5)*(3/2)^(n-5)*1^5.

Compare to the fifth term:
(n choose 5)*(3/2)^(n-5)*1^5 > (n choose 4)*(3/2)^(n-4)*1^4
(n!/(n-5)!5!) > (n!/(n-4)!4!)*3/2
n*(n-1)*(n-2)*(n-3)*(n-4)/5! > n*(n-1)*(n-2)*(n-3)/4! * 3/2
(n-4) > 15/2
n > 11.5

Now compare to the seventh term:
(n choose 5)*(3/2)^(n-5)*1^5 > (n choose 6)*(3/2)^(n-6)*1^6
3/2*6 > (n-6)
9 > n-6
15 > n

The answer is probably 12, 13, 14. Maybe 15 as well (depending on whether you want the sixth term to be strictly greater than the others or not). Verify numerically, as I haven't checked my work.

Answer 2

is it 2.5 or 244.140625