Let $X_{n, i}$ is a triangular arrays of r.v.s
$X_{1, 1}$
$X_{2, 1}$, $X_{2, 2}$
$X_{3, 1}$, $X_{3, 2}$, $X_{3, 3}$
...
$X_{n, 1}$, $X_{n, 2}$, ..., $X_{n, n}$
...
so that each column converges in probability to $0$, i.e., for every fixed $i$, $\{X_{n, i}\}\xrightarrow{p} 0$ (convergence with respect to $n$).
Is it correct that the row-wise average $\dfrac{1}{n}\sum\limits_{i=1}^nX_{n, i}\xrightarrow{p} 0$?
Edit: Each row contains r.v.s which are identically distributed but not independent.
It is not. Consider for instance $X_{n,i}=1$ if $i\ge n/2$, $X_{n,i}=0$ otherwise. Then $$\frac{1}{n}\sum_{i=1}^n X_{n,i} \to \frac{1}{2}.$$ Note that the convergence is actually deterministic in this case (the problem is not really probabilistic in this sense).
Edit: this does not work even if the $(X_{n,i})_{i=1…n}$ are identically distributed. To see this, let $U$ be a r.v. uniformly distributed over $\{1,…,n\}$ and let $X_{n,U}=n$, $X_{n,i}=0$ if $i\neq U$. Then, the $(X_{n,i})_{i=1…n}$ are identically distributed (even exchangeable), $X_{n,i}\to 0$ in probability as $n\to\infty$ (since $\Pr(X_{n,i}=0)=1-1/n$) and yet, for all $n$, $$\frac{1}{n}\sum_{i=1}^n X_{n,i}=1.$$