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Thread: Arithmetic progression

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    Arithmetic progression

    The first and third term of arithmetic progression are x and y respectively.Find the sixteenth term in term of x and y.
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    Quote Originally Posted by mastermin346 View Post
    The first and third term of arithmetic progression are x and y respectively.Find the sixteenth term in term of x and y.
    The values of the arithmetic progression are calculated by

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

    with

    $\displaystyle d = a_{n+1} - a_n$

    In your case $\displaystyle d = \frac{y-x}2$.

    Plug in the values you know:

    $\displaystyle n = 16$
    $\displaystyle a_1 = x$
    $\displaystyle a_3 = y$

    and calculate $\displaystyle a_{16}$.
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    Quote Originally Posted by mastermin346 View Post
    The first and third term of arithmetic progression are x and y respectively.Find the sixteenth term in term of x and y.
    $\displaystyle x + (x+d) + (x+2d) + (x+3d) + ... (x + 15d) $
    where $\displaystyle d $ is the constant difference between each successive term.

    In our case, we are told that the third term is y. Thus, $\displaystyle x+2d = y $ .

    So, $\displaystyle d = \frac{y-x}{2} $


    $\displaystyle a_1 = x $

    $\displaystyle a_2 = x + \frac{y-x}{2} = \frac{2x}{2} + \frac{y-x}{2} =
    \frac{2x+y-x}{2} = \frac{x+y}{2}
    $

    $\displaystyle a_3 = \frac{x+y}{2} + \frac{y-x}{2}= \frac{2y}{2} = y $
    ...
    ...
    ...


    $\displaystyle a_{16} = x + 15d = x + 15\big{(}\frac{y-x}{2}\big{)} $
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    Quote Originally Posted by mastermin346 View Post
    The first and third term of arithmetic progression are x and y respectively.Find the sixteenth term in term of x and y.
    $\displaystyle y = x + 2d \Rightarrow d = \frac{y - x}{2}$. And the first term is x.

    You now have everything necessary to do anything you want, including finding the sixteenth term.
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