The coefficients are....
C(n,8) = (n(n-1)...(n-6)(n-7))/8!
C(n,9) = (n(n-1)...(n-6)(n-7)(n-8))/9!
C(n,10) = (n(n-1)...(n-6)(n-7)(n-8)(n-9))/10!
We need to find the value(s) of n for which the three coefficients are in arithmetic progression -- i.e., for which C(n,9) is the arithmetic mean of C(n,8) and C(n,10).
An interesting problem; but the algebra works out relatively easily....
(n(n-1)...(n-6)(n-7)(n-8))/9! = ((n(n-1)...(n-6)(n-7))/8!+(n(n-1)...(n-6)(n-7)(n-8)(n-9))/10!)/2
Multiply by 2*10! and cancel the common factors n through n-7:
The two solutions are n=14 and n=23.
Check:
For n=14, the coefficients are 1001, 2002, and 3003; 2002 = (1001+3003)/2.
For n=23, the coefficients are 490314, 817190, and 1144066; 817190 = (490314+1144066)/2
DONE!