Question 1151136
<br>
Let the original three numbers be x-6, x, and x+6.<br>
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.<br>
{{{(x-1)/(x-10) = (3x+9)/(x-1)}}}
{{{3x^2-21x-90 = x^2-2x+1}}}
{{{2x^2-19x-91 = 0}}}
{{{(2x+7)(x-13) = 0}}}<br>
{{{x = -7/2}}} or {{{x = 13}}}<br>
Both solutions satisfy the conditions of the problem.<br>
(A) x = -7/2:<br>
The original arithmetic progression is<br>
-19/2, -7/2, 5/2<br>
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:<br>
-27/2, -9/2, -3/2<br>
(B) x = 13:<br>
The original arithmetic progression is<br>
7, 13, 19<br>
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:<br>
3, 12, 48<br>