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:

  1. 13.513.5

  2. 11.2511.25

  3. 10.510.5

  4. 109.5109.5

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)(3, x, y, 9) so that first:

(3,x,y)(3, x, y) is a GP, meaning that there is an rr so that x=3rx = 3r and y=xr=3r2y = xr = 3r^2;

and second:

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

So we have equations in dd and rr.

y=x+d⇝d=y−x=3r2−3r9=x+2d=3r+2(3r2−3r)9=3r+6r2−6r6r2−3r−9=02r2−r−3=(r+1)(2r−3)=0\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=−1r = -1 or 32\dfrac{3}{2}. If we use r=−1r = -1 we won't get positive values, so we go with r=3/2r = 3/2. Then x=3r=92x = 3r = \dfrac{9}{2} and y=3r2=274y = 3r^2 = \dfrac{27}{4}. We wind up with the sequence (3,92,274,9)\left(3, \dfrac{9}{2}, \dfrac{27}{4}, 9\right). The sum of the two middle numbers is 92+274=184+274=454=11.25\dfrac{9}{2} + \dfrac{27}{4} = \dfrac{18}{4} + \dfrac{27}{4} = \dfrac{45}{4} = 11.25.

The answer is (b).