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 value of $m+n$ is $\qquad$

The sequence is:

$\ln \dfrac{a}{b}, \quad \ln \dfrac{a}{b\sqrt{b}}, \quad \ln \dfrac{a}{b^{2}}, \quad \ln \dfrac{a}{b^{2}\sqrt{b}}, \dots$

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The power of $b$ in the denominator for each term is:

$1, \quad \dfrac{3}{2}, \quad 2, \quad \dfrac{5}{2}, \quad \dots$

since $b\sqrt{b} = b^{1} \cdot b^{1/2} = b^{3/2}$, and so on.

These powers increase by $\dfrac{1}{2}$ each time.

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Rewriting each term using log properties:

$\ln \dfrac{a}{b^k} = \ln a - k \ln b$

So each term differs from the previous by $-\dfrac{1}{2}\ln b$, which is constant.

This means the entire sequence is an AP with:

$a_1 = \ln a - \ln b$

$d = -\dfrac{1}{2}\ln b$

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Using the AP sum formula:

$S_{21} = \dfrac{21}{2}\left(2a_1 + 20d\right)$

$= \dfrac{21}{2}\left(2(\ln a - \ln b) + 20\left(-\dfrac{1}{2}\ln b\right)\right)$

$= \dfrac{21}{2}\left(2\ln a - 2\ln b - 10\ln b\right)$

$= \dfrac{21}{2}\left(2\ln a - 12\ln b\right)$

$= 21\ln a - 126\ln b$

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Converting back to a single logarithm:

$S_{21} = \ln a^{21} - \ln b^{126} = \ln \dfrac{a^{21}}{b^{126}}$

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Comparing with $\ln \dfrac{a^m}{b^n}$, we get $m = 21$ and $n = 126$.

$m + n = 21 + 126 = 147$