Question 1151135: Three numbers form a geometric progression. If 4 is subtracted from the third term, then the three numbers will form an arithmetic progression. If, after this, 1 is subtracted from the second and third terms of the progression, then it will again result in a geometric progression. Find these three numbers.
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
There are probably nicer ways to solve this problem.... But this is what I came up with.
The original geometric progression is
a, ar, ar^2
When 4 is subtracted from the third number, the resulting progression
a, ar, ar^2-4
is an arithmetic progression.
Then when 1 is subtracted from the second and third terms, the resulting progression
a, ar-1, ar^2-5
is again a geometric progression.
(1) Using the fact that the last progression is geometric:
 [1]
(2) Using the fact that the second progression is arithmetic:
 [2]
(3) Substituting [2] in [1]....
Both solutions satisfy the conditions of the problem.
(A) r = 3
From [1], a = 1/(2(3)-5) = 1/1 = 1.
The original sequence is
1, 3, 9
When 4 is subtracted from the third number, the resulting sequence is
1, 3, 5
which is an arithmetic progression.
Then when 1 is subtracted from each of the second and third terms, the resulting sequence is
1, 2, 4
which is again a geometric progression.
(B) r = 7
From [1], a = 1/(2(7)-5) = 1/9.
The original sequence is
1/9, 7/9, 49/9
When 4 is subtracted from the third number, the resulting sequence is
1/9, 7/9, 13/9
which is an arithmetic progression.
Then when 1 is subtracted from each of the second and third terms, the resulting sequence is
1/9, -2/9, 4/9
which is again a geometric progression.
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