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Forum URL: http://www.cut-the-knot.org/cgi-bin/dcforum/forumctk.cgi
Forum Name: High school
Topic ID: 380
#0, Geometric series ratios
Posted by ke_45 on Jun-17-08 at 10:59 AM
We have an arithmetic progression the squares of whose 12h, 13th, and 15th terms form a geometric progression. Find all the ratios of this progression.

Assuming

u_n+1 = u_n + d -or- u_n+1 = u_1 + d*n

for the arithmetic progression, and

u_n+1 = u_n * q -or- u_n+1 = u_1 * q^(n-1)

for the geometric one, how do we go about this? (My results don't match with the answer in the book I took this from.)

Please only hint, don't solve!


Thanks
-- KE


#1, RE: Geometric series ratios
Posted by alexb on Jun-17-08 at 11:02 AM
In response to message #0
If the initial arithmetic progrssion is a +i·d, then the condition is

(a + 15d)² / (a + 13d)² = (a + 13d)² / (a + 12d)²

which gives you

(a + 15d)² · (a + 11d)² = (a + 13d)² · (a + 13d)².

There are two ways to get rid of the square. Try both.