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anonymous
 4 years ago
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 !
anonymous
 4 years ago
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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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1a,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 = 8a = c+4/3 hmmm

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1do we have an answer guide to chk with in the end?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.06 , 2 , 2/3 or 16/3 , 8/3 , 4/3

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1those are option i assume ...

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.13c 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 = 8b = 8(c+4/3) \[\frac{(8c\frac{4}{3})\cfrac{9(8c\frac{4}{3})c^2}{9c^2+12c+16}}{1\cfrac{3c}{3c+4}}=8\] would then solve for C :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1wolframalpha simplifies that to: http://www.wolframalpha.com/input/?i=%28%2820%2F3c%29%289c%5E2%2820%2F3c%29%29%2F%289c%5E2%2B12c%2B16%29%29%2F%281%283c%29%2F%283c%2B4%29%298%3D0

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1im sure theres a simpler way to di this, i just cant think of one at the moment ....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have another method . In general, the equation for 'find the number of term' is \[ar ^{n1}\] , 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.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i like that pathway of thinkg alot better :) kudos

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[a+ar=8\] \[ar\frac{4}{3}=ar^2;\ ar=ar^2+\frac 43\] \[a+ar^2+\frac 43=\frac {24}3\] \[ar^2+a+\frac {20}3=0\] \[a(r^2+1)=\frac {20}3\] \[r^2+1=\frac {20}{3a}\] \[r^2=\frac {20}{3a}1\] \[r=\sqrt{\frac {20}{3a}1}\] that looks fun :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[r=\sqrt{\frac {20}{3a}1}\]perpetuated a  lol
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