Question 1099029
<br>Let the first term be x, and let the common ratio of the geometric sequence be r.  Then the first three terms of the sequence are
{{{matrix(1,3,x,xr,xr^2)}}}<br>
The last three of the four terms form an arithmetic sequence.  The difference between the second and third terms is {{{x(r^2-r)}}}; so the fourth term is the third term, plus that difference: {{{xr^2 + x(r^2-r) = x(2r^2-r)}}}<br>
So now the means are {{{xr}}} and {{{xr^2}}}, and the extremes are {{{x}}} and {{{x(2r^2+r)}}}<br>
The sum of the means is 24, and the sum of the extremes is 32:
{{{x(r+r^2) = 24}}}
{{{x(2r^2-r+1) = 32}}}<br>
We can divide these two equations one by the other to eliminate x, giving us an equation we can solve for a:<brh
{{{(2r^2-r+1)/(r^2+r) = 32/24 = 4/3}}}
{{{6r^2-3r+3 = 4r^2+4r}}}
{{{2r^2-7r+3 = 0}}}
{{{(2r-1)(r-3) = 0}}}<br>
Somewhat surprisingly -- to me, at least -- it appears we have two potential solutions to the problem.  Let's see if they both work....<br>
(1) If r = 3...<br>
The sequence is x, 3x, 9x, 15x.
x+15x = 32; 16x = 32; x=2.  The sequence is 2, 6, 18, 30.  Yes; that one works.<br>
(2) If r = 1/2...<br>
The sequence is x, x/2, x/4, 0.
x+0 = 32; x = 32.  The sequence is 32, 16, 8, 0.  Yes, that one works also.<br><br>
There are two sequences that satisfy the given conditions:
2, 6, 18, 30   and   32, 16, 8, 0.