the inner loop is additive increase this makes growth as $\mathcal{O}(n)$
the outer loop is multiplicative decrease, this makes shrinkage as $\mathcal{O}(\log_2 (n) )$
both are in serial connection, therefore $\text{Time Complexity} = \mathcal{O}\left( n\log_2(n) \right)$
answer = option B
the recurrence relation that governs the value of $j$ is given as:
$$\begin{align*} j_m &= j_{m-1} + m \\ a_n &= a_{n-1} + n \qquad \text{; rewriting the relation} \end{align*}$$
such a recurrence relation is non-homogeneous recurrence relation.
whose solution is in the form $$\begin{align*} a_n &= a_n^p + a_n^h\\ &= \text{Particular Solution} + \text{Homogeneous Solution} \end{align*}$$
for homogeneous solution we can see it directly as $r=1$ so $a_n^h = c_h$ where $c_h$ is a constant.
the $F(n)$ here is a linear function whose trial solution will be in the form $d_0+d_1 n$
combining which gives us that the inner loop is $\mathcal{O}(n)$
