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:

- 13½
- 11¼
- 10½
- 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* = 3*r*, *y* = *xr* = 3*r*^{2}.

*x*, *y*, 9 is an AP: *y* = *x* + *d*, 9 = *y* + *d* = *x* + 2*d*.

So we have equations in *d* and *r*.

*y* = *x* + *d* → *d* = *y – x* = 3*r*^{2} – 3*r*

9 = *x* + 2*d* = 3*r* + 2(3*r*^{2} – 3*r)*

9 = 3*r* + 6*r*^{2} – 6*r*

6*r*^{2} – 3*r* – 9 = 0

2*r*^{2} – *r* – 3 = 0

(*r* + 1)(2*r* -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* = 3*r* = 9/2, *y* = 3*r*^{2} = 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).