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Thread: Sequences, Series, and Probability

  1. #1
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    Sequences, Series, and Probability

    Find a formula for an for the arithmetic sequence.

    a2 = 93, a6 = 65


    any help would be greatly appreciated. thank you.
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  2. #2
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    Quote Originally Posted by thepride View Post
    Find a formula for an for the arithmetic sequence.

    a2 = 93, a6 = 65


    any help would be greatly appreciated. thank you.
    what is the formula for the nth term of an arithmetic sequence?
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    Quote Originally Posted by skeeter View Post
    what is the formula for the nth term of an arithmetic sequence?
    im pretty sure its an = dn + c

    but what do i do??? i missed classed and my professor doesnt work on the weekends?
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  4. #4
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    Quote Originally Posted by thepride View Post
    im pretty sure its an = dn + c
    better look it up and make sure ... you'll need it to solve this problem.
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    Quote Originally Posted by skeeter View Post
    better look it up and make sure ... you'll need it to solve this problem.
    ok i found this an= dn + c this is linear form

    and i also found this an= a1 +(n -1)d this is alternative form. hope this helps.
    Last edited by flexus; Sep 26th 2009 at 12:29 PM. Reason: add
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  6. #6
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    Quote Originally Posted by flexus View Post
    ok i found this an= dn + c this is linear form

    and i also found this an= a1 +(n -1)d this is alternative form. hope this helps.
    two people involved w/ this problem?

    $\displaystyle a_n = a_1 + (n-1)d$


    $\displaystyle a_2 = 93$ ...

    *** $\displaystyle 93 = a_1 + (2-1)d$

    $\displaystyle a_6 = 65$ ...

    *** $\displaystyle 65 = a_1 + (6-1)d$

    you have two equations (***) with two unknowns ... solve the system for $\displaystyle d$ and $\displaystyle a_1$
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  7. #7
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    Quote Originally Posted by skeeter View Post
    two people involved w/ this problem?

    $\displaystyle a_n = a_1 + (n-1)d$


    $\displaystyle a_2 = 93$ ...

    *** $\displaystyle 93 = a_1 + (2-1)d$

    $\displaystyle a_6 = 65$ ...

    *** $\displaystyle 65 = a_1 + (6-1)d$

    you have two equations (***) with two unknowns ... solve the system for $\displaystyle d$ and $\displaystyle a_1$
    no i was just helping him/her out.
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  8. #8
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    Binomial Theorem

    good luck trying to solve that one thepride. i have no clue.
    Last edited by flexus; Sep 26th 2009 at 01:12 PM. Reason: wrong post
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  9. #9
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    These are two linear equations so should be relatively simple to solve. Already I can see that the common difference will be negative because $\displaystyle a_2 > a_6$
    Hint:

    $\displaystyle
    65 = a_1 + (6-1)d
    $

    is equal to

    $\displaystyle a_1 = 65-5d$

    -------------

    I get

    $\displaystyle d = -14$

    $\displaystyle a_1 =135$
    Last edited by e^(i*pi); Sep 26th 2009 at 01:57 PM. Reason: Put in final answer to aid the OP
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  10. #10
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    Quote Originally Posted by e^(i*pi) View Post
    These are two linear equations so should be relatively simple to solve. Already I can see that the common difference will be negative because $\displaystyle a_2 > a_6$
    Hint:

    $\displaystyle
    65 = a_1 + (6-1)d
    $

    is equal to

    $\displaystyle a_1 = 65-5d$

    -------------

    I get

    $\displaystyle d = -14$

    $\displaystyle a_1 =135$
    thats not right i check the back of the book and the answer is An= -7n+107

    can someone help me understand this problem???
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  11. #11
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    Quote Originally Posted by thepride View Post
    thats not right i check the back of the book and the answer is An= -7n+107

    can someone help me understand this problem???

    HI

    we know that

    $\displaystyle a_2=93 \Rightarrow a+d=93 $ ---- 1

    and

    $\displaystyle a_6=65 \Rightarrow a+5d=65$ ----2

    Then solving the simultaneous equatino would give

    $\displaystyle d=-7$ and $\displaystyle a=100$

    $\displaystyle T_n=100+(n-1)(-7)$

    =the given answer .
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