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Math Help - Arithmetic Progression

  1. #1
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    Arithmetic Progression

    Hello! I just learnt arithmetic progression and I am confused about all the formula, so I would appreciate if someone showed me an example using the question below...Thank you very much!!

    Q: In an arithmetic progression, the sum of the first 4 terms is 22 and the sum of the first 5 terms is 35. Find the hundredth term.

    Thank you!
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by Tangera View Post
    Hello! I just learnt arithmetic progression and I am confused about all the formula, so I would appreciate if someone showed me an example using the question below...Thank you very much!!

    Q: In an arithmetic progression, the sum of the first 4 terms is 22 and the sum of the first 5 terms is 35. Find the hundredth term.

    Thank you!
    a_1 ~~=~~ a_1

    a_2 ~~=~~ a_1 + x

    a_3 ~~=~~ a_2 + x ~~=~~ a_1 +2x

    a_4 ~~=~~ a_3 + x ~~=~~ a_1 +3x

    a_5 ~~=~~ a_4 + x ~~=~~ a_1 +4x


    You were told the sum of the first 4 terms is 22, so:
    a_1 + (a_1+x) + (a_1+2x) + (a_1 +3x) ~~~~=~~~~ 4*a_1 +6x ~~~~=~~~~ 22

    You were told the sum of the first 5 terms is 35, so you can solve for the 5th term:
    35-22 = 13, so ~~~~a_5 ~~~~=~~~~ a_1+4x ~~~~=~~~~ 13



    Now you have two equations and two variables.

    4*a_1 + 6x = 22
    a_1 + 4x = 13~~~~~\Rightarrow a_1 = 13 - 4x

    Substitute this into the first equation
    4*(13-4x) + 6x = 22

    52-10x = 22

    x = 3

    And substitute this back into either equation
    a_1 + 4(3) = 13
    a_1 = 1



    So now you have your values:
    a_1 = 1
    a_2 = 1+3
    a_3 = 1+3*2
    a_4 = 1+3*3
    a_5 = 1+3*4

    The pattern should be rather obvious: a_n ~~~~=~~~~ 1+3(n-1) ~~~~=~~~~ 3n-2

    So a_{100} ~~~~=~~~~ 3*100-2~~~~=~~~~ 298
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  3. #3
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    Hello, Tangera!

    In an arithmetic progression, the sum of the first 4 terms is 22
    and the sum of the first 5 terms is 35. .Find the hundredth term.
    We are expected to know these formulas . . .

    \text{Given the first term, }a\text{, and the common difference, }d:

    . . \text{The }n^{th}\text{ term is: }\;a_n \;=\;a + (n-1)d

    . . \text{The sum of the first }n\text{ terms: }\;n \:=\:\frac{n}{2}[2a + (n-1)d]


    We are told: . S_4\,=\,22
    . . So we have: . \frac{4}{2}[2a + 3d] \:=\:22 \quad\Rightarrow\quad 2a + 3d \:=\:11\;\;{\color{blue}[1]}

    We are told: . S_5\,=\,35
    . . So we have: . \frac{5}{2}[2a + 4d] \:=\:35 \quad\Rightarrow\quad 2a + 4d \:=\:14\;\;{\color{blue}[2]}

    Subtract [1] from [2]: . \boxed{d \:=\:3}

    Substitute into [1]: . 2a + 3(3) \:=\:11\quad\Rightarrow\quad\boxed{ a \:=\:1}


    Therefore: . a_{100} \;=\;a + 99d \;=\;1 + 99(3) \quad\Rightarrow\quad\boxed{ a_{100}\;=\;298}

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