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Math Help - GMAT questions

  1. #1
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    GMAT questions

    Hi everyone,

    I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?

    Here's a question:

    In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?

    I already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.
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  2. #2
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    Quote Originally Posted by crazirani View Post
    Hi everyone,

    I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?

    Here's a question:

    In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?

    I already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.
    t(n)=t(n-1)-3=t(n-2)-3-3=...=t(n-(n-1))-3(n-1),

    but t(n-(n-1))=t(1), so:

    t(n)=t(1)-3(n-1),

    so now we have:

    t(n)=23-3n+3=26-3n

    If t(n)=4, we have:

    -4=26-3n,

    or:

    3n=26+4=30,

    so n=10.

    RonL
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  3. #3
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    Quote Originally Posted by crazirani View Post
    Hi everyone,

    I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?

    Here's a question:

    In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?

    I already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.
    Method 2.

    t(n) clearly declines linearly as n increases, so put:

    t(n)=kn+c.

    When t(1)=23, and t(2)=20, so we have:

    k+c=23
    2k+c=20,

    so subtracting the first from the second gives:

    (2k+c)-(k+c)=20-23

    or:

    k=-3.

    Substitute this back into the first equation to ge:

    c=26,

    so:

    t(n)=26-3n,

    and the rest is as before.

    RonL
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  4. #4
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    Quote Originally Posted by crazirani View Post
    Hi everyone,

    I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?

    Here's a question:

    In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?
    Here is a linear algebra approach.

    You can solve the recurrence relation.

    t(n)-t(n-1)=-3
    And initial condition is t(1)=23.

    First we solve the homogenous equation,
    t(n)-t(n-1)=0
    The charachteristic equation is,
    k-1=0
    k=1.
    Thus, the general solution is,
    t(n)=C(1)^n=C
    Where C is a constant.

    To find a specfic solution to,
    t(n)-t(n-1)=-3
    We look for a solution in the form,
    t(n)=an
    t(n-1)=a(n-1)=an-a
    Substitute,
    an-an+a=-3
    a=-3
    Thus,
    The specific solution is,
    t(n)=-3n
    Thus, the general solution is the sum of particular and general solution to homogenous equation,
    t(n)=-3n+C
    Intitial conditions,
    t(1)=-3(1)+C=+23
    C=26
    Thus,
    t(n)=-3n+26

    And we need to solve,
    -3n+26=-4
    -3n=-30
    n=-10
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  5. #5
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    Thanks Ron, I'll have to study your responses very carefully and try to do some more problems to really understand what's going on.
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  6. #6
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    Thanks TPH.
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  7. #7
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    Quote Originally Posted by ThePerfectHacker View Post
    Here is a linear algebra approach.

    You can solve the recurrence relation.

    t(n)-t(n-1)=-3
    And initial condition is t(1)=23.

    First we solve the homogenous equation,
    t(n)-t(n-1)=0
    The charachteristic equation is,
    k-1=0
    k=1.
    Thus, the general solution is,
    t(n)=C(1)^n=C
    Where C is a constant.

    To find a specfic solution to,
    t(n)-t(n-1)=-3
    We look for a solution in the form,
    t(n)=an
    t(n-1)=a(n-1)=an-a
    Substitute,
    an-an+a=-3
    a=-3
    Thus,
    The specific solution is,
    t(n)=-3n
    Thus, the general solution is the sum of particular and general solution to homogenous equation,
    t(n)=-3n+C
    Intitial conditions,
    t(1)=-3(1)+C=+23
    C=26
    Thus,
    t(n)=-3n+26

    And we need to solve,
    -3n+26=-4
    -3n=-30
    n=-10
    Interesting. (I learned all my Linear Algebra in a Quantum Physics class, so there are broad gaps in my knowledge.) I couldn't help but notice how similar this solution pattern is to solving a similar looking ODE. Is there any relationship between the two that you know of?

    -Dan
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  8. #8
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    Quote Originally Posted by topsquark View Post
    Is there any relationship between the two that you know of?
    Eigenvalues.
    The chrachteristic polynomial that we get from,
    \det (k\bold{I}-\bold{A})=0.

    To add....
    If,
    A\bold{x}=\bold{b}
    Is a linear system.
    With n equations and m unknowns.

    And then all solutions are,
    x_0+\mbox{Nullspace}(A)
    Where \mbox{Nullspace}(A) is the basis for A\bold{x}=\bold{0} meaning "the general solution".

    See the same pattern is here as well.
    Last edited by ThePerfectHacker; January 18th 2007 at 01:27 PM.
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  9. #9
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    Quote Originally Posted by topsquark View Post
    Interesting. (I learned all my Linear Algebra in a Quantum Physics class, so there are broad gaps in my knowledge.) I couldn't help but notice how similar this solution pattern is to solving a similar looking ODE. Is there any relationship between the two that you know of?
    Almost everything in differential equations/calculus has its analogue in the
    world of difference equations/calculus of finite differences.

    Integration by parts has an analogous (two forms actually) summation by
    parts formula, which I wasted a lot of time playing with once upon a time.

    RonL
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