Expert: Paul Klarreich - 5/22/2008

Question
Hi Paul, I'm not sure where to start with this question: Three numbers form an arithmetic sequence, the common difference is 11.  If the first number is decreased by 6, the second is decreased by 1, and the third number is doubled, the resulting numbers form a geometric sequence.  Determine the three numbers that form the geometric sequence.

Answer
Questioner:   Patrick
Category:  Advanced Math
Private:  No
 
Subject:  Sequences
Question:  Hi Paul, I'm not sure where to start with this question: Three numbers form an arithmetic sequence, the common difference is 11.  If the first number is decreased by 6, the second is decreased by 1, and the third number is doubled, the resulting numbers form a geometric sequence.  Determine the three numbers that form the geometric sequence.
...................................
Hi, Patrick,

Start with definitions:

Three numbers in a.s. can be written:

a, a + d, a + 2d.

It the c.d. is 11, write;

a, a + 11, a + 22.

If the first number is decreased by 6,

write  a - 6

the second is decreased by 1,

write a + 10

and the third number is doubled  

write  2a + 44

....................
If three numbers form a g.s. then there is a common ratio:

Second   Third
------ = -----
First    Second


a + 10   2a + 44
------ = -------
a - 6    a + 10


a^2 + 20a + 100 = 2a^2 + 32a - 264

a^2 + 12a - 364 = 0


(a + 26)(a - 14) = 0

a = -26, a = 14

If a = - 26, the a.s. is -26,-15,-4 and the g.s. would be:

-32, -16, -8, so your common ratio is 2.


If a = 14, the a.s. is 14, 25, 36 and the g.s. would be:

8, 24, 72, so your common ratio is 3.

Both solutions seem valid.