Recursive Sequence $a_n = \frac{1}{2} (a_{n-1} + 5) $

It might help to solve the recurrence relation

Define $b_n\equiv a_n+c$ and find the value of $c$ that makes $b_n$ a geometric sequence

$$a_n = \frac{1}{2} (a_{n-1} + 5)\implies b_n-c = \frac{1}{2} (b_{n-1}-c + 5)$$

$$\implies b_n = \frac{1}{2} b_{n-1}-\frac c2 + \frac 52 +c$$ $$\implies b_n = \frac{1}{2} b_{n-1} + \frac {c+5}2$$

SO if $c=-5$ we have $ b_n = \frac{1}{2} b_{n-1} $ which is a geometric sequence having solution

$$ b_n = b_1 \left( \frac 12\right) ^{n-1}$$

Where $b_1=a_1-5 = 3-5 = -2$

So $$ a_n = b_n + 5 = 5 -2 \left( \frac 12\right) ^{n-1}=5 - \left( \frac 12\right) ^{n-2} $$

You can view the given recurrent sequence in this way: The $(n+1)$-th term is the average of $n$-th term and $5$. From this it is relatively easy to see that the distance from $5$ will be halved in each step. And this can be verified by a direct computation: $$5-a_{n+1} = 5- \frac{a_n+5}2 = \frac{5-a_n}2.$$

As already explained in another answer, you get from this that $(5-a_n)$ is a geometric sequence which tends to zero for $n\to\infty$.

Another way to visualize this is notice that you have $a_{n+1}=f(a_n)$ for $$f(x)=\frac{5+x}2.$$ This gives the following picture

In this case, the function is very simple, so maybe this is more complicated than other solution. But it is perhaps useful to know that if the function $f(x)$ fulfills some conditions, then the sequence determined by $a_{n+1}=f(a_n)$ will always converge to the solution of $f(x)=x$. Look up Banach fixed-point theorem for more information.

But even without relying on some theorem, if the function is simple enough and you can plot it, you can see from the graph whether these iterations converge to $f(x)=x$. You can find picture illustrating this for various functions here. Some of them are taken from related posts:

• "Why do I always get 1 when I keep hitting the square root button on my calculator?"
• Iteration for solving $x=g(x)$.

And I will also add that such diagrams are called cobweb plots.