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.