Question 1151135
<br>
There are probably nicer ways to solve this problem....  But this is what I came up with.<br>
The original geometric progression is<br>
a, ar, ar^2<br>
When 4 is subtracted from the third number, the resulting progression<br>
a, ar, ar^2-4<br>
is an arithmetic progression.<br>
Then when 1 is subtracted from the second and third terms, the resulting progression<br>
a, ar-1, ar^2-5<br>
is again a geometric progression.<br>
(1) Using the fact that the last progression is geometric:<br>
{{{(ar^2-5)/(ar-1) = (ar-1)/a}}}
{{{a^2r^2-2ar+1 = a^2r^2-5a}}}
{{{-2ar+1 = -5a}}}
{{{1 = 2ar-5a}}}
{{{1 = a(2r-5)}}}
{{{a = 1/(2r-5)}}} [1]<br>
(2) Using the fact that the second progression is arithmetic:<br>
{{{(ar^2-4)-ar = ar-a}}}
{{{ar^2-2ar+a = 4}}} [2]<br>
(3) Substituting [2] in [1]....<br>
{{{r^2/(2r-5)-2r/(2r-5)+1/(2r-5) = 4}}}
{{{r^2-2r+1 = 4(2r-5)}}}
{{{r^2-2r+1 = 8r-20}}}
{{{r^2-10r+21 = 0}}}
{{{(r-3)(r-7) = 0}}}<br>
Both solutions satisfy the conditions of the problem.<br>
(A) r = 3<br>
From [1], a = 1/(2(3)-5) = 1/1 = 1.<br>
The original sequence is<br>
1, 3, 9<br>
When 4 is subtracted from the third number, the resulting sequence is<br>
1, 3, 5<br>
which is an arithmetic progression.<br>
Then when 1 is subtracted from each of the second and third terms, the resulting sequence is<br>
1, 2, 4<br>
which is again a geometric progression.<br>
(B) r = 7<br>
From [1], a = 1/(2(7)-5) = 1/9.<br>
The original sequence is<br>
1/9, 7/9, 49/9<br>
When 4 is subtracted from the third number, the resulting sequence is<br>
1/9, 7/9, 13/9<br>
which is an arithmetic progression.<br>
Then when 1 is subtracted from each of the second and third terms, the resulting sequence is<br>
1/9, -2/9, 4/9<br>
which is again a geometric progression.<br>