1972 AHSME Problems/Problem 16
Problem 16
There are two positive numbers that may be inserted between and 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
Solution 1
Let be first and be second. We can then get equations based on our knowledge: and . We then get from our first equation and we substitute that into the second to get which simplifies to which be and . Then . Their sum is . ~lopkiloinm
Solution 2
Like in Solution 1, let be the first number and be the second. From the problem, we have that
for some common difference and common ratio . We can use these to obtain the following:
(from , , and the definition of an arithmetic sequence)
(from and )
Solving for and substituting into , we find that or . Moving every term to the right side and factoring, we get that . Because and must be positive, the common ratio cannot be , so we ascertain that the common ratio is . From here, we find that , and . Then their sum is .
~ cxsmi