Somewhere in between...

The format of the answer allows one to solve this problem without actually calculating the limit, as every term of this sequence $a_n$ is of the form $a_1^pa_2^q$ where $p+q=1$ . We can prove this by induction.

Base cases: $a_1=a_1^1a_2^0\\a_2=a_1^0a_2^1$ Inductive step:

Assume $\begin{aligned}a_{n-1}&=a_1^wa_2^x,&w+x&=1\\a_n&=a_1^ya_2^z,&y+z&=1\end{aligned}$ Then $a_{n+1}=a_1^{(w+y)/2}a_2^{(x+z)/2}\\\frac{w+y}{2}+\frac{x+z}{2}=1$ Therefore, $l+m+n=2$

I would modify the last part of the problem to:

If $\displaystyle\lim_{n\rightarrow \infty}a_{ n }=l(a_{ 2 })^{ p }(a_{ 1 })^{ q }$ ,

then $\displaystyle l+p+q$ is

$\boxed {Solution}$

We can find the next five terms of this sequence:

$\displaystyle a_{3}=\sqrt {a_{2}a_{1}} =[a_{2}a_{1}]^{\frac{1}{2}}$

$\displaystyle a_{4}=\sqrt {a_{3}a_{2}} =\sqrt {\displaystyle(\sqrt {a_{2}a_{1} })(a_{ 2 }}) =\sqrt[\displaystyle4]{(a_{2})^{3}a_{1}} =\left[\displaystyle (a_{2})^{3}a_{1}\right]^{\frac{1}{4}}$

$\displaystyle a_{5}=\sqrt {a_{4}a_{3}} =\sqrt {\left(\sqrt[\displaystyle 4]{(a_{2})^{3}a_{1}}\right)(\sqrt {a_{2}a_{1}})} =\sqrt[\displaystyle 8]{(a_{2})^{5}(a_{1})^{3}} =\left[(a_{2})^{5}(a_{1})^{3}\right]^{\frac{1}{8}}$

$\displaystyle a_{6}=\sqrt {a_{5}a_{4}} =\sqrt {\left(\sqrt[\displaystyle 8]{(a_{2})^{5}(a_{1})^{3}}\right)\left(\sqrt[\displaystyle4]{(a_{2})^{3}a_{1}}\right)} =\sqrt[\displaystyle 16]{(a_{2})^{11}(a_{1})^{5}} =\left[(a_{2})^{11}(a_{1})^{5}\right]^{\frac{1}{16}}$

$\displaystyle a_{7}=\sqrt {a_{6}a_{5}} =\sqrt {\left(\sqrt[\displaystyle 16]{(a_{2})^{11}(a_{1})^{5}}\right)\left(\sqrt[\displaystyle 8]{(a_{2})^{5}(a_{1})^{3}}\right)} =\sqrt[\displaystyle 32]{(a_{2})^{21}(a_{1})^{11}} =\left[(a_{2})^{21}(a_{1})^{11}\right]^{\frac{1}{32}}$

We can see that (not too easy, it took me a long time) $\displaystyle a_{n} = \left[(a_{2})^{\frac{2^{n-1}+(-1)^{n}}{3}} \cdot (a_{1})^{\frac{2^{n-2}+(-1)^{n-1}}{3}}\right]^{\frac{1}{2^{n-2}}}$

It can be simplified to:

$\displaystyle a_{n} = (a_{2})^{\frac{2}{3}+\frac{4}{3} \left(-\frac{1}{2}\right)^{n}} \cdot (a_{1})^{ \frac{1}{3}-\frac{4}{3} \left(-\frac{1}{2}\right)^{n}}$

Now, applying the limit:

$\displaystyle\lim_{n\rightarrow \infty}a_{ n }= (a_{2})^{\frac{2}{3}} \cdot (a_{1})^{\frac{1}{3}}$

Therefore, $\displaystyle l=1, p= \frac{2}{3}, q= \frac{1}{3}$ and $l+p+q =\boxed {2}$

Firstly we will take $ln({a}_{n})={b}_{n}$ and our recurrence becomes :

${b}_{n+2}=\frac{{b}_{n+1}+{b}_{n}}{2}$

Simplifying the recurrence :

$2{b}_{n+2}-{b}_{n+1}-{b}_{n}=0$

Now it's a linear recurrence relation and hence we will find the charecteristic polynomial of it and it comes out to be :

$2{x}^{2}-x-1=0$

It's roots are ${1},\frac{-1}{2}$

Hence the general solution is of the form :

$\displaystyle {b}_{n}={c}_{1}+{c}_{2}{(\frac{-1}{2})}^{n}$

Putting the initial values ${b}_{1}$ and ${b}_{2}$ we get :

$\displaystyle {b}_{n}=\frac{{b}_{1}+2{b}_{2}}{3} + \frac{2({b}_{1}-{b}_{2})}{3}{(\frac{-1}{2})}^{n-1}$

Applying limit $n$ tending to infinity we get :

$\displaystyle \lim _{ n\rightarrow \infty }{ { b }_{ n } } = \frac{{b}_{1}+2{b}_{2}}{3}$

Reverting back to our substitution we get :

$\displaystyle \lim _{ n\rightarrow \infty }{ { ln(a }_{ n }) } = \frac{ln({a}_{1})+2ln({a}_{2})}{3}$

Hence $\displaystyle \lim _{ n\rightarrow \infty }{ { a }_{ n } } = {a}_{1}^{\frac{1}{3}}{a}_{2}^{\frac{2}{3}}$

To learn more about the method of solving linear recurrences by charecteristic polynomial method refer to Recurrence relations .

From the equation for $a_{n+1}$ we derive,

$\prod_{i=3}^{n}a_i =\sqrt{(a_2 a_1)( a_3 a_2)( a_4 a_3) ... (a_{n-2} a_{n-1}) }$

$\prod_{i=3}^{n}a_i =\sqrt{ a_1 a_2^2 a_3^2 a_4^2 ... a_{n-2}^2 a_{n-1} }$

$\prod_{i=3}^{n}a_i = a_2 a_3 a_4 ... a_{n-2} \sqrt{ a_1 a_{n-1}}$

$a_{n-1} a_n = a_2 \sqrt{ a_1 a_{n-1}}$

$\lim_{n \to \infty} {(a_{n-1} a_n) } = \lim_{n \to \infty} (a_2 \sqrt{ a_1 a_{n-1}})$

Since $\lim_{n \to \infty} {(a_{n-1}) } = \lim_{n \to \infty} {(a_{n})}$

$l^2 a_1^{2 p} a_2^{2 q} = a_2 \sqrt{a_1} \sqrt{l a_1^p a_2^q}$ using p,q for m,n resp.

Squaring yields $l^4 a_1^{4 p} a_2^{4 q} = l a_1^{p+1} a_2^{q+2}$

So l=1, p= $\frac{1}{3}$ and q = $\frac{2} {3}$ and l + p + q = $\boxed{2}$