SOLUTION: Find a sequence of four numbers. The first of which is six and the fourth of which is 16, if the first three numbers form an arithmetic sequence and the last three numbers form a g

Question 978756: Find a sequence of four numbers. The first of which is six and the fourth of which is 16, if the first three numbers form an arithmetic sequence and the last three numbers form a geomtrice sequence
Found 2 solutions by richwmiller, htmentor:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
6,6+x,6+2x,16
r=(6+2x)/6+x),
16=(6+2x)/(6+x)*(6+2x)
two sequences
r=-4 x=-5
6,1,-4,16
and
r=4/3 x=3
6,9,12,16

Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
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

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