How find this value:$\dfrac{f(a_{1})+f(a_{3})}{f(a_{2})}$ 
Let $f(x)=\ln{x}-\dfrac{1}{x}+3$, and $a_{i}>0,i=1,2,3$, such that $a_{3}:a_{2}:a_{1}=e^2:e:1$.
Suppose $f(a_{1})+f(a_{2})+f(a_{3})=\dfrac{e^5-e^2}{1-e}$; what is the value of $\dfrac{f(a_{1})+f(a_{3})}{f(a_{2})}$? (where $e=2.718\cdots, \ln{x}=\log_{e}{x}$)
 A: $\dfrac{e^2+1}{e}\quad $ B:$\dfrac{e^2+3}{e+1}\quad$ C :$\dfrac{e^2+5}{e+2}\quad$ D :$\dfrac{e^3+e+2}{e^2+1}$


this problem have nice solution? Thank you
my idea:$a_{3}=e^2*a_{1},a_{2}=e*a_{1}$
since $f(a_{1})+f(a_{2})+f(a_{3})=3\ln{a_{1}}+3-\dfrac{1}{a_{1}}\left(1+\dfrac{1}{e}+\dfrac{1}{e^2}\right)+9=\dfrac{e^5-e^2}{1-e}$,
$$\Longrightarrow 3\ln{a_{1}}=\dfrac{1}{a_{1}}\left(1+\dfrac{1}{e}+\dfrac{1}{e^2}\right)+\dfrac{e^5-e^2}{1-e}-12$$
then find the value
$$\dfrac{f(a_{1})+f(a_{3})}{f(a_{2})}=\dfrac{2\ln{a_{1}}+8-\dfrac{1}{a_{1}}(1+\dfrac{1}{e^2})}{\ln{a_{1}}-\dfrac{1}{a_{1}}\dfrac{1}{e}+4}=\dfrac{\dfrac{1}{3a_{1}}\left(\dfrac{2}{e}-1-\dfrac{1}{e^2}\right)+\dfrac{2}{3}\dfrac{e^5-e^2}{1-e}}{-\dfrac{1}{3a_{1}}\left(\dfrac{2}{e}-1-\dfrac{1}{e^2}\right)+\dfrac{1}{3}\dfrac{e^5-e^2}{1-e}}$$
then we must find the $a_{1}$,Now I think follow is very ugly.someone have  nice methods? Thank you
oh, I have solution:let $a_{2}=x$, It's easy have
$$3\ln{x}+9=(e^2+e+1)\left(\dfrac{1}{ex}-e^2\right)$$
so let
$$F(x)=3\ln{x}+9-(e^2+e+1)\left(\dfrac{1}{ex}-e^2\right),x>0$$
so
$F'(x)=\dfrac{3}{x}+\dfrac{e^2+e+1}{e}\dfrac{1}{x^2}>0$
and we have $F(e^{-3})=0$
so $a_{2}=e^{-3}$
and follwing is very easy
A: Hint$$f=\ln e^3x -\dfrac1x$$
$$f(a_1)+f(a_2)+f(a_3)=3ln(ae^2)-\dfrac1a\bigg[1+\dfrac1e+\dfrac1{e^2}\bigg]$$ 
$$=3\ln(ae^2)-\dfrac1a\bigg[\dfrac{e^3-1}{e-1}\bigg]$$
where $a$ is 1st term 
of the $GP:(a1,a2,a3)$
Now , $$f(a_1)+f(a_2)+f(a_3)=-\dfrac{e^5-e^2}{e-1}$$
Comparing both the equations.$$ln(ae^2)=0\ ;\ \dfrac1a\bigg[\dfrac{e^3-1}{e-1}\bigg]=\dfrac{e^5-e^2}{e-1}$$
So, $$a=\dfrac1{e^2}$$
Now we can proceed use this value for a
