5 numbers in AP, GP and HP. The Question: Consider 5 numbers $a_1,a_2,a_3,a_4,a_5$ such that $a_1,a_2,a_3$ are in AP, $a_2,a_3,a_4$ are in GP and $a_1,a_4,a_5$ are in HP. Then find whether $\ln a_1,\ln a_3, \ln a_5$ are in AP,GP or HP.
My attempt: I tried to form a specific example taking $a_1=0,a_2=1,a_3=2,a_4=4,a_5=8$ . Now calculating the logarithms of $a_1,a_3$ and $a_5$ I couldn't find any specific pattern.
I also tried to write down the terms using the common difference and common ratio and tried substituting them but couldn't find any relation among the needed quantities. Please help, thanks in advance. 
EDIT- $a_3, a_4,a_5$ must be in HP, not $a_1,a_4,a_5$ as pointed out by Harsh and farruhota. My textbook had the question wrong.
 A: They are in AP( arithmetic progression). 
Simply using the conditions for the numbers to be in respective progressions. 
Since $a_1 , a_2 ,a_3$ are in AP so 
$$ 2a_2 = a_1 +a_3 $$ ------(i)
Since $ a_2 ,a_3, a_4 $are in GP
$$ (a_3)^2 = a_2 a_4 $$--------(ii)
Since  $ a_3 , a_4 ,a_5 $ are in HP
So $$ \frac{2}{a_4}  = \frac{1}{a_3} + \frac{1}{a_5}$$----------(iii)
Using eq ii
$$ a_2= \frac{(a_3)^2}{a_4} $$
Using equation iii 
$$ a_4 = \frac{2a_3 a_5}{a_3 + a_5} $$
$$ a_2= \frac{(a_3)^2}{a_4} $$
$$ a_2= \frac{(a_3)^2}{ \frac{2a_3 a_5}{a_3 + a_5}} $$
$$ a_2= \frac{a_3(a_3 +a_5)}{2a_5} $$
Substituting the value of$a_2 $ in eq ii:
$$ \frac{a_3(a_3+a_5)}{a_5} = a_1 + a_3 $$
$$ (a_3)^2 +a_3 a_5 = a_1 a_5 + a_3 a_5 $$
$$ (a_3)^2 = a_1 a_5 $$
Taking ln both sides, 
and using its properties we get:
$$ 2ln(a_3) = ln(a_1)+ln(a_5) $$
This is clearly the condition for numbers to be in AP.
So $lna_1 ,lna_3 , lna_5$ they are in AP
A: For the symmetry, it must be $AP: a_1,a_2,a_3, GP: a_2,a_3,a_4, HP: a_{\color{red}3},a_4,a_5$. Hence:
$$AP:a_1,a_2,a_3 \iff a_2-d,a_2,a_2+d\\
GP: a_2,a_3,a_4 \iff  a_2,a_2r,a_2r^2\\
a_2+d=a_2r \Rightarrow a_2=\frac{d}{r-1}\\
HP: a_3,a_4,a_5 \iff \frac1{a_3}+\frac1{a_5}=\frac2{a_4} \Rightarrow \\
a_5=\frac{a_3a_4}{2a_3-a_4}=\frac{a_2r\cdot a_2r^2}{2a_2r-a_2r^2}=\frac{a_2r^2}{2-r}=\frac{dr^2}{(2-r)(r-1)}.$$
The given is AP:
$$\ln a_1+\ln a_5=2\ln a_3 \Rightarrow \\
a_5=\frac{a_3^2}{a_1}=\frac{(a_2r)^2}{a_2-d}=\frac{(\frac{dr}{r-1})^2}{\frac{d}{r-1}-d}= \frac{dr^2}{(2-r)(r-1)}.$$
