nth term的maths問題

a) common difference = 13-16 = -3
nth term = 16-3(n-1)

b) common ratio = (-6)/2 = -3
nth term = 2*(-3)^(n-1)

c) T(3)-T(2) = 9/5-(-6/13) = 147/65
T(2)-T(1) = -6/13-1/8 = -61/104
∴No common difference
∴It is not A.S.
T(3)/T(2) = (9/5)/(-6/13) = -39/10
T(2)/T(1) = (-6/13)/(1/8) = -48/13
∴No common ratio
∴It is not G.S.

a) a = 16
There exists a common difference d = 13-16 = 10-13 = 7-10 = -3
The series is an AP.
Hence, the nth term T(n) = a+(n-1)d = 16-3(n-1) = 19 - 3n

b) a = 2
There exists a common ratio r = (-6)/2 = 18/(-6) = (-54)/18 = -3
The series is a GP.
Hence, the nth term T(n) = a r^(n-1) = 2*(-3)^(n-1)

c) There is no common difference nor common ratio found,
however, the series can be rewritten as:
2/16, -6/13, 18/10, -54/7 ....
It can be observed that each term is a combination of the terms in (a) and (b).
The numerator comes from (b) and the denumerator comes from (a).
Hence, the nth term T(n) = [2*(-3)^(n-1)] / (19 - 3n)