Question 978756
The sequence of numbers is 6, b, c, 16
The first three numbers form an arithmetic sequence:
b = 6 + d and c = 6 + 2d, where d = the common difference
The last three numbers form a geometric sequence.  The first number of the sequence is b:
c = br and 16 = br^2, where r = the common ratio
r = c/b = (6+2d)/(6+d)
16 = br^2 = (6+d)((6+2d)/(6+d))^2 = (6+2d)^2/(6+d)
Solve for d:
(36+24d+4d^2)/(6+d) = 16
36 + 24d + 4d^2 = 96 + 16d
4d^2 + 8d - 60 = 0
d^2 + 2d - 15 = 0
(d-3)(d+5) = 0
This gives two possible answers, d=3, d=-5
So the two sequences are 6, 9, 12, 16 or 6, 1, -4, 16
The first sequence has common ratio c/b = 12/9 = 4/3
The second sequence has common ratio c/b = -4/1 = -4