The second term of geometric sequence exceeds the third term by 4/3 and the sum of the first two terms is 8. Find the possible values of the first three terms. Thank for the solution !

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The second term of geometric sequence exceeds the third term by 4/3 and the sum of the first two terms is 8. Find the possible values of the first three terms. Thank for the solution !

Mathematics
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a,b,c are what we have to work with; and its geometric ... b = c+ 4/3 b c --- = ----- ; common ratio a c+4/3 a+b = 8 b = 8-a = c+4/3 hmmm
do we have an answer guide to chk with in the end?
6 , 2 , 2/3 or 16/3 , 8/3 , 4/3

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those are option i assume ...
3c b ----- = --- 3c+4 a 3ca = b(3c+4) hard to tell which route to take .... \[\frac{a-\cfrac{9ac^2}{9c^2+12c+16}}{1-\cfrac{3c}{3c+4}}=8\] a = 8-b = 8-(c+4/3) \[\frac{(8-c-\frac{4}{3})-\cfrac{9(8-c-\frac{4}{3})c^2}{9c^2+12c+16}}{1-\cfrac{3c}{3c+4}}=8\] would then solve for C :)
wolframalpha simplifies that to: http://www.wolframalpha.com/input/?i=%28%2820%2F3-c%29-%289c%5E2%2820%2F3-c%29%29%2F%289c%5E2%2B12c%2B16%29%29%2F%281-%283c%29%2F%283c%2B4%29%29-8%3D0
im sure theres a simpler way to di this, i just cant think of one at the moment ....
I have another method . In general, the equation for 'find the number of term' is \[ar ^{n-1}\] , where a is the first term, r is common difference, n is number of term. So a, ar, ar^2 , what we need to work with. From question, we can two equation, 1) a+ar =8 ; 2) ar = ar^2 + 4/3 .Use simultaneous equation, you are able to solve.
r is common ratio*
i like that pathway of thinkg alot better :) kudos
\[a+ar=8\] \[ar-\frac{4}{3}=ar^2;\ ar=ar^2+\frac 43\] \[a+ar^2+\frac 43=\frac {24}3\] \[ar^2+a+\frac {-20}3=0\] \[a(r^2+1)=-\frac {20}3\] \[r^2+1=-\frac {20}{3a}\] \[r^2=-\frac {20}{3a}-1\] \[r=\sqrt{-\frac {20}{3a}-1}\] that looks fun :)
\[r=\sqrt{\frac {20}{3a}-1}\]perpetuated a - lol

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