<PRE><B>Question 1191284</B>
&lt;br&gt;
Let a and r be the first term and common ratio.  The given information tells us&lt;br&gt;
(1) a+ar+ar^2+ar^3+...=8
(2) ar+ar^2=3&lt;br&gt;
We can write the infinite sum using the formula for the sum of an infinite geometric series:&lt;br&gt;
(3) {{{a/(1-r)=8}}}&lt;br&gt;
Subtracting (2) from (1), we get&lt;br&gt;
(4) a+ar^3+ar^4+...=5&lt;br&gt;
Use the formula for the sum of an infinite geometric series again on all but the first term:&lt;br&gt;
(5) {{{a+(ar^3/(1-r))=5}}}&lt;br&gt;
Substitute (3) into (5):&lt;br&gt;
{{{a+8r^3=5}}}
{{{8r^3=5-a}}}
{{{r^3=(5-a)/8}}}
(6) {{{r=((5-a)/8)^(1/3)}}}&lt;br&gt;
Substitute (6) into (3):&lt;br&gt;
{{{a/(1-((5-a)/8)^(1/3))=8}}}&lt;br&gt;
{{{a=8(1-((5-a)/8)^(1/3))=8-8((5-a)^(1/3))/2=8-4(5-a)^(1/3)}}}
{{{4(5-a)^(1/3)=8-a}}}
{{{(5-a)^(1/3)=2-a/4}}}&lt;br&gt;
At this point, it is easily seen that a=4 will work.&lt;br&gt;
Algebraic or numerical methods, or a graphing calculator, can be used to show that a=4 is the ONLY solution to that equation, so it will lead us to the unique answer for the common ratio r.&lt;br&gt;
{{{r=((5-a)/8)^(1/3)=(1/8)^(1/3)=1/2}}}&lt;br&gt;
ANSWER: the common ratio of the series is 1/2&lt;br&gt;
CHECK:&lt;br&gt;
The sequence has first term 4 and common ratio 1/2: 4, 2, 1, 1/2, 1/4, ...
The infinite sum is {{{4/(1-1/2)=8}}}
The sum of the second and third terms is 2+1=3&lt;br&gt;
NOTE:
The solution a=4 and r=1/2 can be found by playing with numbers on the original problem.  Since the problem asked for the sum of all the possible values of the common ratio, we had to do some algebra to show that there are no other possible values for the common ratio.</PRE>