A species has an initial population $4^{10}.$ At the end of first day, the population increases by $50\%$. At the end of second day, it decreases by the same percentage. If the process continues in the same pattern, the number of days for the population to reach $3^{10}$ is

Let $a_n$be the population at the end of $nth$ day, then

$\begin{array}{l}{a}_n=\left\{\begin{array}{c}{4}^{10}{\left(\frac{3}{2}\right)}^{n/2}{\left(\frac{1}{2}\right)}^{n/2}\text{ if }n\text{ is even }\\ 4^{10}{\left(\frac{3}{2}\right)}^{(n+1)/2}{\left(\frac{1}{2}\right)}^{(n-1)/2}\text{ is }n\text{ is odd }\end{array}\right.\\ =\left\{\begin{array}{c}{4}^{10}(3/4{)}^{n/2}\text{ if }n\text{ is even }\\ 4^{10}{\left(\frac{3}{4}\right)}^{(n-1)/2}\left(\frac{3}{2}\right)\text{ if }n\text{ is odd }\end{array}\right.\end{array}$

As all answers are even natural numbers we set $a_n=3^{10}$ assuming $n$ to be even.
$\Rightarrow 4^{10}{\left(\frac{3}{4}\right)}^{n/2}=3^{10}\Rightarrow n=20$