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Thread: 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!
    $\displaystyle a_1 ~~=~~ a_1$

    $\displaystyle a_2 ~~=~~ a_1 + x$

    $\displaystyle a_3 ~~=~~ a_2 + x ~~=~~ a_1 +2x$

    $\displaystyle a_4 ~~=~~ a_3 + x ~~=~~ a_1 +3x$

    $\displaystyle a_5 ~~=~~ a_4 + x ~~=~~ a_1 +4x$


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

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



    Now you have two equations and two variables.

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

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

    $\displaystyle 52-10x = 22$

    $\displaystyle x = 3$

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



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

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

    So $\displaystyle a_{100} ~~~~=~~~~ 3*100-2~~~~=~~~~ 298$
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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 . . .

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

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

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


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

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

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

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


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

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