Question 1151136: Three numbers are in an arithmetic progression with a common difference of 6. If 4 is subtracted from the first number, 1 is subtracted from the second number, and the third number is first decreased by 3 and then multiplied by 3, then the resulting three numbers form a geometric progression. Find the original three numbers.
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
Let the original three numbers be x-6, x, and x+6.
Then (x-6)-4 = x-10, x-1, and 3((x+6)-3) = 3x+9 form a geometric progression, which means there is a common ratio between the terms.
 or 
Both solutions satisfy the conditions of the problem.
(A) x = -7/2:
The original arithmetic progression is
-19/2, -7/2, 5/2
When 4 is subtracted from the first term, 1 is subtracted from the second term, and the third term is decreased by 3 and then multiplied by 3, the resulting numbers form a geometric progression:
-27/2, -9/2, -3/2
(B) x = 13:
The original arithmetic progression is
7, 13, 19
When 4 is subtracted from the first term, 1 is subtracted from the second term, and the third term is decreased by 3 and then multiplied by 3, the resulting numbers form a geometric progression:
3, 12, 48
|
|
|
