If the sum of the first \(21\) terms of the sequence: \(\ln \frac{a}{b}, \ln \frac{a}{b \sqrt{b}}, \ln \frac{a}{b^{2}}, \ln \frac{a}{b^{2} \sqrt{b}}, \ldots\) is \(\ln \frac{a^{m}}{b^{n}}\), then…

The sum has $m+n=147$. Each of the $21$ terms contributes one $\ln a$, and the exponents of $b$ run from $1$ to $11$ in steps of $\frac12$, so their sum is $\frac{21(1+11)}{2}=126$. The value $126$ leaves out the contribution $m=21$.