A particle moving in a straight line having acceleration which varies with velocity as $a=-kv^n (n\ne1,2)$. Here k is a constant. For what value of $n$ the average velocity of the particle averaged over the time, till it stops is one third of the initial velocity?
$$\newcommand{\b}[1]{\left(#1\right)} \newcommand{\d}{{\rm d}} \newcommand{\f}{\frac} \newcommand{\s}{\sqrt} \newcommand{\t}{\text} \newcommand{\u}{\underbrace} \newcommand{\r}{\implies}$$ My Solution: $$a=\f{\d v}{\d t}=-kv^n\r\int_{v_0}^{v} v^{-n}\d v=-\int_0^t k\d t\r \f1{1-n}\b{v^{1-n}-v_0^{1-n}}=-kt\\v=\b{v_0^{1-n}+(n-1)kt}^{\f1{1-n}}$$ Now: $$\langle v\rangle =\f{\int_0^{\infty} v\d t}{\int_0^{\infty}\d t}=\lim_{\tau\to\infty}\f1{\tau}\int_0^{\tau}\b{v_0^{1-n}+(n-1)kt}^{\f1{1-n}}\d t\\ =\lim_{\tau\to\infty}\f1{\tau}\b{\f1{\b{\f1{1-n}+1}(n-1)k}\b{v_0^{1-n}+(n-1)kt}^{\f1{1-n}+1}}_0^{\tau}=\f{v_0}3\t{(given)}\\ \lim_{\tau\to\infty}\f1{\tau}\b{\f1{(n-2)k}\b{\b{v_0^{1-n}+(n-1)k\tau}^{\f1{1-n}+1}-v_0^{2-n}}}=\f{v_0}3$$ This becomes a little messy here as I am not able to solve further.
I have tried just a few minutes ago L'Hospital too: $$\langle v\rangle=\f1{(n-2)k}\b{\b{\f1{1-n}+1}\b{v_0^{1-n}+(n-1)k\tau}^{\f1{1-n}}(n-1)k}$$ But it also doesn't seem to work.
When $n>1$ your expectation $\langle v\rangle$ from L'Hopitals rule, goes to zero as $\tau \to \infty$. (I have not checked your algebra here, but that agrees with my own derivation for the $n>1$ case)
You are asked about the average velocity from t=0 untill the particle stops. My reading is that the particle stops after a finite time, in which case it stops at: $$ t_{\rm{stop}}=\frac{-v_0^{1-n}}{(n-1)k} $$ Which to be a finite positive time requires that $n<1$.
Now you need to find $n<1$ such that that the average velocity over the interval $[0,t_{\rm{stop}}]$ be $v_0/3$