If the sum of the first 21 terms of the sequence: $$\ln \frac{a}{b}, \ln \frac{a}{b\sqrt{b}}, \ln \frac{a}{b^2}, \ln \frac{a}{b^2\sqrt{b}}, ........$$ is $$\ln \frac{a^m}{b^n} $$, then the value of $$m + n$$ is

$$a_1=\ln\left(\dfrac{a}{b}\right),\ a_2=\ln\left(\dfrac{a}{b^{\frac{3}{2}}}\right)$$. Thus, the common difference will be = $$a_2-a_1$$

=> $$d=\ln\left(\dfrac{a}{b^{\frac{3}{2}}}\right)-\ln\left(\dfrac{a}{b}\right)=\ln\left(\dfrac{a\times b}{a\times b^{\frac{3}{2}}}\right)=-\dfrac{1}{2}\ln\left(b\right)$$

The, sum of first n terms is given by = $$\dfrac{n}{2}\left[2a+\left(n-1\right)d\right]$$

Here, n = 21; thus, the sum of the first 21 terms = $$\dfrac{21}{2}\left[2a+20d\right]=21\left[a+10d\right]$$

=> $$S_{21}=21\left[\ln\left(\dfrac{a}{b}\right)+10\left(-\dfrac{1}{2}\right)\ln b\right]$$

=> $$S_{21}=21\left[\ln a-\ln b-5\ln b\right]$$

=> $$S_{21}=21\left[\ln a-6\ln b\right]$$

=> $$S_{21}=21\ln\left(\dfrac{a}{b^6}\right)$$

=> $$S_{21}=\ln\left(\dfrac{a^{21}}{b^{126}}\right)$$

It is given that the sum of first 21 terms = $$\ln\left(\dfrac{a^m}{b^n}\right)$$, therefore m = 21 and n = 126.

Thus, $$m+n=21+126=147$$