Originally Posted by
mastermin346 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 $
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$\displaystyle a_{16} = x + 15d = x + 15\big{(}\frac{y-x}{2}\big{)} $