Problem: There are two positive numbers that may be inserted between 3 and 9 such that the first three are in geometric progression while the last three are in arithmetic progression. The sum of those two positive numbers is
Answer Choices:
A. 1321​
B. 1141​
C. 1021​
D. 10
E. 921​
Solution:
Let 3,x,y be the first three numbers. Then 3x​=xy​,x2=3y. The last three numbers are x,y,9 to that y−x=9−y,x+9=2y. Eliminate y getting 2x2−3x−27=0. Factoring, (x+3)(2x−9)=0, x=431​,y=3x2​=643​.∴x+y=1141​.