The third term of a sequence is 2(2/3) and the sixth term is 9. Find the first six terms when the sequence is; (a) arithmetic (b) geometric
a) for arithmetic:
t3 = a+2d = 4/3
t6 = a+5d = 9
subtract them
3d = 23/3
d = 23/9
then a+2(23/9) = 4/3
a = -34/9
first 6 terms are:
-34/9, -11/9 , 4/3 , 35/9 , 58/9 , 9
well, that checked out ok
b) for geometric
t3 = ar^2 = 4/3
t6 = ar^5 = 9
divide them
r^3 = 27/4 = 54/8 , so I can take a nice cube root of the bottom
r = (3/2)(2^(1/3)
then a(9/4)(2^(2/3) = 4/3
a = (16/27) 2^(-2/3)
messy subs , you try them