SOLUTION: 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

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

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