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kryton1212

  • one year ago

a and b are +ve integers. a,-4,b form a geometric sequence, and -4,b,a form an arithmetric sequence. (a) Find the value of ab. (b) Find the values of a and b. (c) (i) Find the sum to infinity of the geometric sequence a,-4,b,... . (ii) Find the sum to infinity of all the terms that are +ve in the geometric sequence a,-4,b.

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  1. kryton1212
    • one year ago
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    @hartnn

  2. hartnn
    • one year ago
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    if x,y,z form an geometric sequence, then \(y^2 =xz\) so, what about a,-4 and b ?

  3. kryton1212
    • one year ago
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    ab=(-4)^2=16 why?

  4. kryton1212
    • one year ago
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    it is also fixed?

  5. hartnn
    • one year ago
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    i can get you a simple explanation here... each next term of geometric sequence is MULTIPLIED by the common ratio 'r' so, in general, terms are a,ar,ar^2,ar^3..... when we have just 3 terms, we have a, ar, ar^2 for this (1st term * 3rd term) = (2nd term)^2 (you can check)

  6. kryton1212
    • one year ago
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    okay, i understand

  7. hartnn
    • one year ago
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    or in other words, they form a proportion, \(\dfrac{1st \: term}{2nd \: term}=\dfrac{2nd \: term}{3rd\: term}\)

  8. hartnn
    • one year ago
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    16 is correct :) now if the terms are in arithmetic sequence, then 2* 2nd term = 1st term + 3rd term use this...

  9. kryton1212
    • one year ago
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    wow, i like the formulae

  10. hartnn
    • one year ago
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    or if you don't want to remember the formula, you should know that difference between the terms is constant in arithmetic sequence so, difference between b and -4 = difference between a and b b - (-4) = a-b you'll get same equation :)

  11. kryton1212
    • one year ago
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    yup, got it

  12. hartnn
    • one year ago
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    tell me what u get for a and b when you get it....2 equations, 2 unknowns...

  13. kryton1212
    • one year ago
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    -4+d=b b+d=a ar=-4 ar^2=b are the equations correct?

  14. hartnn
    • one year ago
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    ar = -4 ? how ?

  15. kryton1212
    • one year ago
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    i don't know, i just guess.....it's written couple of days ago. i forgot how...

  16. hartnn
    • one year ago
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    forget about the 'd' and 'r' now we have 2 equations in a and b ab = 16 and from b-(-4) = a-b, we have 2b =a-4 you got these ? how..

  17. kryton1212
    • one year ago
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    uh, you said this moment ago. got it. a=8 , b=2

  18. hartnn
    • one year ago
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    those values are correct. so b) is done right ? now c) part can you find the common ratio in 8,-4,2,....?

  19. kryton1212
    • one year ago
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    *-2

  20. hartnn
    • one year ago
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    b=2 was correct....

  21. kryton1212
    • one year ago
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    common ratio is -2

  22. hartnn
    • one year ago
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    ohh! yes, r= -2 :) first term = a1 = 8 SUm to infinite is just \(\large \dfrac{a_1}{r-1}\)

  23. kryton1212
    • one year ago
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    what? -8/3?

  24. hartnn
    • one year ago
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    ohh....the r is not -2 :P

  25. kryton1212
    • one year ago
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    -1/2 ?

  26. hartnn
    • one year ago
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    yes. r = 2nd term / 1st term = 3rd trm/ 2nd term =.... = -4/8 = -1/2

  27. kryton1212
    • one year ago
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    -16/3 ...

  28. hartnn
    • one year ago
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    since r is LESS than 1, we use \(\huge S_\infty = \dfrac{a_1}{1-r}\)

  29. kryton1212
    • one year ago
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    lol

  30. hartnn
    • one year ago
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    yes, there are 2 formula, depending on r less than 1 or greater than 1

  31. kryton1212
    • one year ago
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    i will remember. I just forgot at that moment..

  32. hartnn
    • one year ago
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    try the last aprt ?

  33. kryton1212
    • one year ago
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    i hate doing +ve / -ve ....

  34. hartnn
    • one year ago
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    remove all the negatives... 8,-4,2,-1,1/2 .... to 8,2,1/2 .... now whats the common ratio r =... ?

  35. kryton1212
    • one year ago
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    oh, i know what's the world happening! r=1/4

  36. hartnn
    • one year ago
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    yessss!

  37. hartnn
    • one year ago
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    and since 1/4 is less than 1 you use a/ (1-r)

  38. kryton1212
    • one year ago
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    32/3

  39. kryton1212
    • one year ago
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    million thanks

  40. hartnn
    • one year ago
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    \(\huge \color{red}\checkmark\)

  41. hartnn
    • one year ago
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    welcome ^_^

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