$(a_{n,k})$ of numbers is given by $a_{n,1} = \frac 1n$, for $n = 1, 2, \dots , a_{n,k+1} = a_{n-1,k} - a_{n,k}$, for $1 \leq k \leq n - 1$. You need $a_{0,k}$for definition

First, we shall show that the numbers in the $n$-th row are precisely $$\frac{1}{n\binom{n-1}{0}}, \frac{1}{n\binom{n-1}{1}},\cdots,\frac{1}{n\binom{n-1}{n-1}}$$This can be shown easily by induction. $n=1,2,3,4$ can be directly verified as shown in the question statement. We may now show $$\frac{1}{n\binom{n-1}{k-1}}-\frac{1}{(n+1)\binom{n}{k-1}}=\frac{1}{(n+1)\binom{n}{k}}$$Now the harmonic mean of the elements of the $n$-th row is equal to $$\frac{n}{n\binom{n-1}{0}+n\binom{n-1}{1}+\cdots+n\binom{n-1}{n-1}}=\frac{1}{2^{n-1}}$$Hence the harmonic mean of the 1985th row is $\frac{1}{2^{1984}}$.