2870.21 – A Mixed Geometric and Arithmetic Sequence

Find the 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:

13.5
11.25
10.5
109.5

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 also.

Solution

We seek the sequence: $(3, x, y, 9)$ so that first:

$(3, x, y)$ is a GP, meaning that there is an $r$ so that $x = 3r$ and $y = xr = 3r^2$ ;

and second:

$(x, y, 9)$ is an AP, meaning that there is a $d$ such that $y = x + d$ and $9 = y + d = x + 2d.$

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

\begin{aligned} y &=& x + d \leadsto d = y - x = 3r^2 - 3r, \\ 9 &=& x + 2d = 3r + 2(3r^2 - 3r), \\ 9 &=& 3r + 6r^2 - 6r, \\ 6r^2 - 3r - 9 &=& 0, \\ 2r^2 - r - 3 &=& (r + 1)(2r -3) = 0 \end{aligned}

So, $r = -1 \text{ or } 3/2$ . If we use $r = -1$ we won't get positive values, so we go with $r = 3/2$ . Then $x = 3r = 9/2$ and $y = 3r^2 = 27/4.$ We wind up with the sequence (3, 9/2, 27/4, 9). The sum of the two middle numbers is 9/2 + 27/4 = 18/4 + 27/4 = 45/4 = 11.25.

The answer is (b).