Arithmetic progression

**mastermin346** · January 5th 2010, 11:03 PM

The first and third term of arithmetic progression are x and y respectively.Find the sixteenth term in term of x and y.

**earboth** · January 6th 2010, 02:06 AM

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.

The values of the arithmetic progression are calculated by

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

with

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

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

Plug in the values you know:

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

and calculate $a_{16}$ .

**abender** · January 6th 2010, 02:09 AM

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.

$x + (x+d) + (x+2d) + (x+3d) + ... (x + 15d)$

where $d$ is the constant difference between each successive term.

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

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

$a_1 = x$

$a_2 = x + \frac{y-x}{2} = \frac{2x}{2} + \frac{y-x}{2} =<br /> \frac{2x+y-x}{2} = \frac{x+y}{2}<br />$

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

$a_{16} = x + 15d = x + 15\big{(}\frac{y-x}{2}\big{)}$

**mr fantastic** · January 6th 2010, 02:10 AM

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.

$y = x + 2d \Rightarrow d = \frac{y - x}{2}$ . 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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