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    hi guys!

    i've tried question 23 multiple times with attempts to work out r, but i cannot get a numerical value for it as all my equations for r = have two other unknowns (a and d) within them.

    how can a question like this be solved? thank you Name:  image.png
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    (Original post by ashaxo99)
    hi guys!

    i've tried question 23 multiple times with attempts to work out r, but i cannot get a numerical value for it as all my equations for r = have two other unknowns (a and d) within them.

    how can a question like this be solved? thank you Name:  image.png
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    Please post your working - perhaps your most recent attempt
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    I have devised a solution. Can you tell me what the correct answer is and if I'm right, I'll explain it.
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    (Original post by Kevin De Bruyne)
    Please post your working - perhaps your most recent attempt
    my apologies, i should've included this in my main post my working is really messy and probably won't make a lot of sense at all, but i've basically been trying to solve for r, but not with much luck! Name:  image.jpg
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    (Original post by thekidwhogames)
    I have devised a solution. Can you tell me what the correct answer is and if I'm right, I'll explain it.
    thank you for taking the time to do so! the correct answer is D) 1/4
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    (Original post by ashaxo99)
    thank you for taking the time to do so! the correct answer is D) 1/4
    ar = a+4d (1)
    ar^2 = a+5d (2)

    Subtract (1) from (2) to get d in terms of a and r.

    Substitute for d in (1).
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    (Original post by ashaxo99)
    thank you for taking the time to do so! the correct answer is D) 1/4
    This was quite a nice problem - is it for A level (if so it's quite challenging).

    Anyhow, my answer was correct. Here's how I did it:

    Spoiler:
    Show

    Given firstly all the terms are different (note this).

    We know:

    - both first terms are the same
    - ar = a+4d
    - ar^2 = a+5d (I)

    ar=a+4d
    ar^2=a+5d

    Add d to the first equation yields ar+d=a+5d and that's equal to the same thus the expressions must obviously be equal. That is, ar+d=ar^2.

    Then by simple algebra and factorising we get d=ar(r-1)

    Let's put this back into (I); ar^2 = a+5ar(r-a)

    So we know ar^2 = a+5ar^2 - 5ar

    Putting it all to one side gives us 4ar^2-5ar+a=0

    Factoring gives us (4ar-a)(r-1) = 0

    So r can be 1 however that makes all the terms the same hence this solution for is rejected.

    4ar-a=0 --> 4ar=a --> 4r=1 --> r=1/4

    Hence r=1/4 and thus answer is (D).

    Lovely question.
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    (Original post by tiny hobbit)
    ar = a+4d (1)
    ar^2 = a+5d (2)

    Subtract (1) from (2) to get d in terms of a and r.

    Substitute for d in (1).
    thank you very much!
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    (Original post by thekidwhogames)
    This was quite a nice problem - is it for A level (if so it's quite challenging).

    Anyhow, my answer was correct. Here's how I did it:

    Spoiler:
    Show


    Given firstly all the terms are different (note this).

    We know:

    - both first terms are the same
    - ar = a+4d
    - ar^2 = a+5d (I)

    ar=a+4d
    ar^2=a+5d

    Add d to the first equation yields ar+d=a+5d and that's equal to the same thus the expressions must obviously be equal. That is, ar+d=ar^2.

    Then by simple algebra and factorising we get d=ar(r-1)

    Let's put this back into (I); ar^2 = a+5ar(r-a)

    So we know ar^2 = a+5ar^2 - 5ar

    Putting it all to one side gives us 4ar^2-5ar+a=0

    Factoring gives us (4ar-a)(r-1) = 0

    So r can be 1 however that makes all the terms the same hence this solution for is rejected.

    4ar-a=0 --> 4ar=a --> 4r=1 --> r=1/4

    Hence r=1/4 and thus answer is (D).

    Lovely question.

    ah, what a stunning question indeed

    thank you so much for your extremely comprehensive reply! that was super helpful the question is from a cambridge entrance exam for economics! last year's, actually! once again, thank you for taking the time to devise and explain your solution
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    What is this? Is it the TMUA or something else?
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    (Original post by Kaiylar)
    What is this? Is it the TMUA or something else?
    it's from the economics admissions assessment for Cambridge, although the maths questions on it are extremely similar to the TMUA
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    (Original post by ashaxo99)
    it's from the economics admissions assessment for Cambridge, although the maths questions on it are extremely similar to the TMUA
    Ah right.
    Posted on the TSR App. Download from Apple or Google Play
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    (Original post by ashaxo99)
    ah, what a stunning question indeed

    thank you so much for your extremely comprehensive reply! that was super helpful the question is from a cambridge entrance exam for economics! last year's, actually! once again, thank you for taking the time to devise and explain your solution
    No problem, it was good practice for my Maths skills too. Good luck!
 
 
 
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