# Arithmetic progression

• Jan 5th 2010, 11:03 PM
mastermin346
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.
• Jan 6th 2010, 02:06 AM
earboth
Quote:

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

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

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 AM
abender
Quote:

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

$\displaystyle a_1 = x$

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

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

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

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.

$\displaystyle y = x + 2d \Rightarrow d = \frac{y - x}{2}$. And the first term is x.

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