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Math Help - Really hard word problem

  1. #1
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    Really hard word problem

    In a financial arrangement, you are promised $900 the first day and each day after that you will receive 75% of the previous day's amount. When one day's amount drops below $1, you stop getting paid from that day on. What day is the first day you receive no payment and what is your total income? Use a formula for the nth partial sum of a geometric sequence.
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  2. #2
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    You should know the formulas dealing with geometric sequences and series. First you need to find the term where a_n < 1. You have your initial value and common ratio so you can just plug the info into the sequence formula. Then once you know the ending term use the geometric series formula to solve for the sum.

    You know what formulas I mean right?
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  3. #3
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    Quote Originally Posted by Jameson View Post
    You should know the formulas dealing with geometric sequences and series. First you need to find the term where a_n < 1. You have your initial value and common ratio so you can just plug the info into the sequence formula. Then once you know the ending term use the geometric series formula to solve for the sum.

    You know what formulas I mean right?
    yea i know what you mean. but it just really hard when i do word problems. like usually different problems ask me they already give me a1 and r and i just cant figure out the terms. i just need help. like i know the form is An=A1(r)^(n-1) but i just dont see what you see.
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  4. #4
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    Quote Originally Posted by thepride View Post
    yea i know what you mean. but it just really hard when i do word problems. like usually different problems ask me they already give me a1 and r and i just cant figure out the terms. i just need help. like i know the form is An=A1(r)^(n-1) but i just dont see what you see.
    It's ok. I'll walk you through it.

    You have the right formula. All you need to do is figure out a_1 and r. a_1 is the starting value so this should be easy to figure out since there are very few numbers. Now for the ratio, you know it's decreasing each time by 75%, but written in decimal form the ratio is .75. You want to find for what n you get a_n is 1 or less than one. So let a_n=1, plug in the other info I just wrote about and solve for n.

    Does that make more sense?
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  5. #5
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    Quote Originally Posted by Jameson View Post
    It's ok. I'll walk you through it.

    You have the right formula. All you need to do is figure out a_1 and r. a_1 is the starting value so this should be easy to figure out since there are very few numbers. Now for the ratio, you know it's decreasing each time by 75%, but written in decimal form the ratio is .75. You want to find for what n you get a_n is 1 or less than one. So let a_n=1, plug in the other info I just wrote about and solve for n.

    Does that make more sense?
    ok sorry it took so long for me to get back to ya....this website is godly slow.

    ok this is what i got from what you explain don't laugh if im not anywhere close.

    a1=900 because thats what he gets on the first day correct?
    r= 75%=.75

    an=900*.75^(n-1)
    what do you think??
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  6. #6
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    What more help do you need? Jameson told you, "You want to find for what n you get a_n is 1 or less than one". That is, you want to find n so that 900(.75)^n\le 1 or, equivalently, (.75)^n\le \frac{1}{900}= 0.00111.... You can find n by using logarithms or just by calculating (.75)^n for different n. (Start around n= 20.)
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  7. #7
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    i dont know if i did this right but this is what i did but it not the correct answer as the back of the book.

    900(.75)^20 less than or equal to 1
    i got 2.85 less than or equal to 1

    did i do this right.....what is the correct answer?
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