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Math Help - Geometric Sequences 1

  1. #1
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    Post Geometric Sequences 1

    A ball is dropped from a height of 3 m. After each bounce it rises to 75% of its previous height.

    a) State t1 and r: t1 = 3 m r = 0.75

    b) Write the general term for the sequence:
    tn = (t1)(r^n-1)
    tn = (3)(0.75^n-1)

    c) What height does the ball reach after the 6th bounce? (I am not sure how to do this question)
    This was my attempt: I suppose you solve for tn and you plug in 6 into n...
    tn = (3)(0.75^6-1)
    tn = (3)(0.75^5)
    tn = (3)(.2373046875)
    tn = .7119140625

    My answer was wrong, the answer is 53.39 cm.

    d) After how many bounces will the ball reach a height of approx. 40 cm?
    (I cannot solve this one either)
    My attempt: I plugged in 40 cm into tn and try to solve for n:
    40 = (3)(0.75^n-1)
    13.3 = 0.75^n-1 I am stuck here... I do not know how to get n alone with algebra....

    The answer is 7 times.

    Could anyone tell me where I went wrong for c) and d) and what I should have done? Thanks!
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  2. #2
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    Re: Geometric Sequences 1

    I'm not sure why you put "n - 1" in the general term. To get your books answer, just multiply 3/4 six times.

    On the second part ... you need a logarithmic calculation. See the attached pdf.
    Attached Files Attached Files
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  3. #3
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    Re: Geometric Sequences 1

    Quote Originally Posted by Mathnood768 View Post
    A ball is dropped from a height of 3 m. After each bounce it rises to 75% of its previous height.

    a) State t1 and r: t1 = 3 m r = 0.75

    b) Write the general term for the sequence:
    tn = (t1)(r^n-1)
    You are numbering the bounces incorrectly. With tn= (t1)(r^(n-1)) n= 1 initially. After the first bound n= 2, after the third bounce, n= 4, etc.
    Another way to handle this is to take the initial height as t0 so that t1 is after the first bounce, etc. Then you can use tn= (t1)(r^n).

    tn = (3)(0.75^n-1)

    c) What height does the ball reach after the 6th bounce? (I am not sure how to do this question)
    This was my attempt: I suppose you solve for tn and you plug in 6 into n...
    tn = (3)(0.75^6-1)
    tn = (3)(0.75^5)
    tn = (3)(.2373046875)
    tn = .7119140625
    After the sixth bounce n= 7 so t7= (3)(0.75^(7-1))= (3)(0.75^6)

    My answer was wrong, the answer is 53.39 cm.

    d) After how many bounces will the ball reach a height of approx. 40 cm?
    (I cannot solve this one either)
    My attempt: I plugged in 40 cm into tn and try to solve for n:
    40 = (3)(0.75^n-1)
    13.3 = 0.75^n-1 I am stuck here... I do not know how to get n alone with algebra....
    40 cm is .4 meters. Your equation should be .4= (3)(0.75^(n-1). Then .4/3= 0.133333= (0.75)^(n-1). Since the "n" is an exponent, take the logarithm of both sides. log(0.133333)= (n- 1) log(0.75) so n= 1+ log(0.133333)/log(0.75). But remember that this will be after the n- 1 bounce.

    The answer is 7 times.

    Could anyone tell me where I went wrong for c) and d) and what I should have done? Thanks!
    Last edited by HallsofIvy; February 27th 2013 at 05:39 AM.
    Thanks from topsquark and Mathnood768
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  4. #4
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    Re: Geometric Sequences 1

    Thanks! This makes sense to me, and I have one more question,

    40 cm is .4 meters. Your equation should be .4= (3)(0.75^(n-1). Then .4/3= 0.133333= (0.75)^(n-1). Since the "n" is an exponent, take the logarithm of both sides. log(0.133333)= (n- 1) log(0.75) so n= 1+ log(0.133333)/log(0.75). But remember that this will be after the n- 1 bounce.

    So, I got 8 in the calculator, should I just minus 1 for the final answer? Am I allowed to do that? And I should do that every time I have a question like this right?
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