Find the limit of the recursive sequence Definition  of  the  sequence :
$$a_1=a;\\a_2=b;\\$$ and $$\ \ \ a_{n+2}={{a_n+a_{n+1}}\over2}$$ for $n\geq 1$.
Find the limit of this  sequence in terms of $a$ and $b$.
Now in this case , taking $\lim_{n\rightarrow \infty}$  on  both sides does not help at all . So what I did was try to find the $n$-th term  in therms of $a$ and $b$ and then take $n$ to $\infty.$ 
So , using the given recursion formula I wrote down like  $7$-$8$  terms, grouped them  in  various  ways  but  still  could  not  get the pattern of the $n$-th term. 
Terribly  sorry  for  the  lack  of  work/context  in this post of mine but  I tried  for like an hour  to figure  things out  but can't really put  those  scribbles  down  here.
Please  give  me  some  hints  as  to  how  to find the answer .  Thank you.  

If you would like to have a closed form for the individual terms of the recurrence, you can use the method of generating functions. This first step places your sequence as the terms of a power series $G(x) = \sum_{n=1}^\infty a_n x^n$, and then you generate a separate representation using the recursion. It's a long process but very powerful. It's real strength comes from combinatorics, but it can solve linear recurrences well.
$$G(x) = \sum_{n=1}^\infty a_n x^n = ax + bx^2 + \sum_{n=3}^\infty a_n x^n$$
$$= ax + bx^2 + \sum_{n=1}^\infty a_{n+2} x^{n+2}$$
$$= ax + bx^2 + \frac x2 \sum_{n=1}^\infty a_{n+1} x^{n+1} + \frac{x^2}2 \sum_{n=1}^\infty a_{n} x^{n}$$
$$= ax + bx^2 + \frac x2 \left(\sum_{n=1}^\infty a_{n} x^{n} - ax\right) + \frac{x^2}2 \sum_{n=1}^\infty a_{n} x^{n}.$$
Thus recognizing $G(x)$ as the series on the right, we have
$$G(x)= ax + bx^2 + \frac x2 \left(G(x) - ax\right) + \frac{x^2}2 G(x).$$ Thus solving for $G(x)$ we find:
$$G(x)(1-\frac{x}{2}-\frac{x^2}{2})=ax + (b-\frac{a}{2})x^2$$
and
$$G(x) = \frac{2ax}{2-x-x^2} + \frac{(2b-a)x^2}{2-x-x^2}.$$
Using partial fraction decomposition we get:
$$G(x) = (2ax + (2b-a)x^2) \left( \frac{1}{6} \frac{1}{1-\left(-\frac x 2\right)} + \frac13 \frac{1}{1-x} \right)$$
Now applying geometric series identities:
$$G(x) = (2ax + (2b-a)x^2) \sum_{n=0}^\infty\left( \frac{1}{6}  \frac{(-1)^n}{2^n} + \frac13\right) x^n $$
Manipulating series:
$$G(x) = \sum_{n=0}^\infty 2a \left( \frac{1}{6}  \frac{(-1)^n}{2^n} + \frac13\right) x^{n+1} + \sum_{n=0}^\infty (2b-a)\left( \frac{1}{6}  \frac{(-1)^n}{2^n} + \frac13\right) x^{n+2}$$
$$=\sum_{n=1}^\infty 2a \left( \frac{1}{6}  \frac{(-1)^{n-1}}{2^{n-1}} + \frac13\right) x^{n} + \sum_{n=2}^\infty (2b-a)\left( \frac{1}{6}  \frac{(-1)^{n-2}}{2^{n-2}} + \frac13\right) x^{n}$$
$$=a + \sum_{n=2}^\infty \left[ 2a \left( \frac{1}{6}  \frac{(-1)^{n-1}}{2^{n-1}} + \frac13\right) + (2b-a)\left( \frac{1}{6}  \frac{(-1)^{n-2}}{2^{n-2}} + \frac13\right) \right] x^n$$
Thus since $G$ is a power series centered about zero, its coefficients are uniquely determined. For $n \ge 2$ we have: $$a_n = \left[ 2a \left( \frac{1}{6}  \frac{(-1)^{n-1}}{2^{n-1}} + \frac13\right) + (2b-a)\left( \frac{1}{6}  \frac{(-1)^{n-2}}{2^{n-2}} + \frac13\right) \right]$$
now as $n \to \infty$ we are left with $L = \lim_{n\to\infty} a_n = 2a \cdot \frac13 + (2b-a) \cdot \frac13 = \frac{1}{3}a + \frac{2}{3} b$.
