# Geometric and Arithmetric Sequences

There are two positive numbers that may be inserted between 3 and 9 such that the first three numbers are in geometric progression while the last three are in arithmetic progression. The sum of those two positive numbers is:

1. 13½
2. 11¼
3. 10½
4. 109½

Note: There are two other numbers that work, but they’re not both positive. If you go about this problem in a suitably erudite fashion, you’ll turn up this alternative solution too.

We seek 3, x, y, 9 so that:

3, x, y is a GP: x = 3r, y = xr = 3r2.
x, y, 9 is an AP: y = x + d, 9 = y + d = x + 2d.

So we have equations in d and r.

y = x + dd = y – x = 3r2 – 3r
9 = x + 2d = 3r + 2(3r2 – 3r)
9 = 3r + 6r2 – 6r
6r2 – 3r – 9 = 0
2r2r – 3 = 0
(r + 1)(2r -3) = 0
r = -1 or 3/2.

If we use r = -1 we won’t get positive values, so we go with r = 3/2.
x = 3r = 9/2, y = 3r2 = 27/4.

So we wind up with 3, 9/2, 27/4, 9.
The sum of the two middle numbers is 9/2 + 27/4 = 18/4 + 27/4 = 45/4 = 11¼. The answer is (b).