# Terms of a sequence

#### LLand314

##### Member
write the terms a1,a2,a3,a4 of the following sequence. an+1=0.4an+330, a0=550

everytime I get 550 for a1 a2 a3 and a4
is that correct or am I doing it wrong.

#### tkhunny

##### Well-known member
MHB Math Helper
550 * 0.4 = 220

220 + 330 = 550

Pretty clear. Why do you doubt?

#### LLand314

##### Member
It just seemed to easy for Calculus thats all.

Thanks

#### MarkFL

Staff member
If we write the inhomogeneous recursions:

$$\displaystyle a_{n+1}=0.4a_{n}+330$$

$$\displaystyle a_{n+2}=0.4a_{n+1}+330$$

We find the homogenous recursion via differencing:

$$\displaystyle a_{n+2}=1.4a_{n+1}-0.4a_{n}$$

whose associated characteristic roots are:

$$\displaystyle r=\frac{2}{5},\,1$$

and hence the closed form is:

$$\displaystyle A_n=k_1+k_2\left(\frac{2}{5} \right)^n$$

Now, using:

$$\displaystyle A_0=A_1=550$$, we may determine:

$$\displaystyle k_1=550,\,k_2=0$$ and so:

$$\displaystyle A_n=550$$

#### chisigma

##### Well-known member
write the terms a1,a2,a3,a4 of the following sequence. an+1=0.4an+330, a0=550

everytime I get 550 for a1 a2 a3 and a4
is that correct or am I doing it wrong.
Using the procedure described in...

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/

... the difference equation can be written as...

$\displaystyle \Delta_{n}= a_{n+1}-a_{n} = 330 - .6\ a_{n} = f(a_{n})$ (1)

... and the function f(*) is represented here... There is only one attractive fixed point in x=550 and, because the linearity of f(*) the stable point is met at the first step...

Kind regards

$\chi$ $\sigma$