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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 !

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  1. amistre64
    • 4 years ago
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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

  2. amistre64
    • 4 years ago
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    do we have an answer guide to chk with in the end?

  3. anonymous
    • 4 years ago
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    6 , 2 , 2/3 or 16/3 , 8/3 , 4/3

  4. amistre64
    • 4 years ago
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    those are option i assume ...

  5. amistre64
    • 4 years ago
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    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 :)

  6. amistre64
    • 4 years ago
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    im sure theres a simpler way to di this, i just cant think of one at the moment ....

  7. anonymous
    • 4 years ago
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    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.

  8. anonymous
    • 4 years ago
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    r is common ratio*

  9. amistre64
    • 4 years ago
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    i like that pathway of thinkg alot better :) kudos

  10. amistre64
    • 4 years ago
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    \[a+ar=8\] \[ar-\frac{4}{3}=ar^2;\ ar=ar^2+\frac 43\] \[a+ar^2+\frac 43=\frac {24}3\] \[ar^2+a+\frac {-20}3=0\] \[a(r^2+1)=-\frac {20}3\] \[r^2+1=-\frac {20}{3a}\] \[r^2=-\frac {20}{3a}-1\] \[r=\sqrt{-\frac {20}{3a}-1}\] that looks fun :)

  11. amistre64
    • 4 years ago
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    \[r=\sqrt{\frac {20}{3a}-1}\]perpetuated a - lol

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