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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- Jan 5th 2010, 11:03 PMmastermin346Arithmetic progression
The first and third term of arithmetic progression are x and y respectively.Find the sixteenth term in term of x and y.

- Jan 6th 2010, 02:06 AMearboth
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 $\displaystyle a_{16}$. - Jan 6th 2010, 02:09 AMabender$\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 $

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$\displaystyle a_{16} = x + 15d = x + 15\big{(}\frac{y-x}{2}\big{)} $ - Jan 6th 2010, 02:10 AMmr fantastic