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    Sorry could you post the attachment again or something as it just says 'attachment not found'
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    (Original post by years101)
    Attachment 618070how do i work out b?

    i need to work the 1st term which is a right?
    in part a you have find difference
    for b multiply that difference with forth term that is given
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    Once you've worked out the common ratio, you could use that times the fourth term to get the fifth term. However, you might need the first term to solve c)
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    For prt a, the common ratio is found by dividing successive terms.
    Part b, multiply 144 by your answer to a
    For part c, The sum of a GP is a(1-r^n)/(1-r). When r is between -1 and 1, the sum to infinity can be found, as r^n becomes negligibly small, so the 1-r^n term can be removed. You will have to find the first term though
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    (Original post by years101)
    to get a do i sub r into ar^4 or any of them and then find the answer and then divide by 3/4^4 ?
    If you're using the fifth term then yes. I'd use the third or fourth (ar^2 and ar^3) because the numbers are simpler and you know the starting point is correct.
    But yes, that is how you find a.
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    Ok as different exam boards use different letters and symbols for formulas this might be a different formula to what your used to
    so n=ar^(k-1), where n=term, a=starting term and k=number of terms, r= common ratio

    a) so 144=ar^(3-1) and 108=ar^(4-1)
    therefore 144=ar^2 and 108=ar^3
    (144/r^2)=a and (108/r^3)=a
    combined (144/r^2)=(108/r^3)
    rearranged, 144r^3=108r^2
    simplified, 144r=108
    r=(108/144)
    r=0.75

    b) as r=0.75, input back into starting equations
    144=a(0.75^2)
    144/0.75^2=a
    a=256
    check
    108=a(0.75^3)
    108/0.75^3=a
    a=256
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    (Original post by Science Finger)
    Ok as different exam boards use different letters and symbols for formulas this might be a different formula to what your used to
    so n=ar^(k-1), where n=term, a=starting term and k=number of terms, r= common ratio

    a) so 144=ar^(3-1) and 108=ar^(4-1)
    therefore 144=ar^2 and 108=ar^3
    (144/r^2)=a and (108/r^3)=a
    combined (144/r^2)=(108/r^3)
    rearranged, 144r^3=108r^2
    simplified, 144r=108
    r=(108/144)
    r=0.75

    b) as r=0.75, input back into starting equations
    144=a(0.75^2)
    144/0.75^2=a
    a=256
    check
    108=a(0.75^3)
    108/0.75^3=a
    a=256
    First thing, you didnt answer b correctly. You found a (the value is correct), bt yo're meant to be finding the fifth term of the progression
    Also, afaik we are not supposed to post worked solutions
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    (Original post by years101)
    so i did this

    ar^3 = a(3/4)^3

    (3/4)^3 is 27/64 so do i divide by 3/4)^3 now?
    Yes, then you stick the numbers in the formula.
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    (Original post by years101)
    i got a = 1
    when i did this:

    a(3/4)^3

    27/64 divided by 3/4^3 = 1

    i think i did it wrong
    r does equal 3/4. the third term of the sequence is 144, so ar^2=144. Does that help?
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    sigh could do with something relatively easy after getting a headache from what I have been doing all day time to utterly mess this up to help my self esteem even more.

    Okay here goes

    if the third term is 144 and the fourth term is 108 then
    remember that AR^n-1

    so the third term is AR^2=144
    and the fourth term is AR^3=108
    part A
    divide the 4th term by the third term AR^3/AR^2=108/144
    the As cancel out and R^3/R^2 just equals R
    so R=108/144=0.75

    B the fifth term would be the fourth term multiplied by the common ration which is 108*0.75 which= 81

    C sum to infinity is A/1-r

    we know that r= 0.75
    so A /0.25
    so now we have to find A well if A(0.75)^2=144 then A(0.75*0.75)=144 so
    A(0.5625)=144

    A=144/0.5625= 256
    sum to infinity is therefore 256/0.25
    which = 1024.
 
 
 
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