Find the sum:
$\frac{a-b}{a+b}+\frac{3a-2b}{a+b}+\frac{5a-3b}{a+b}+\ldots$ to $11$ terms.

(D) The given series is an Arithmetic Progression $(AP)$ with first term $A = \frac{a-b}{a+b}$.
The common difference $D$ is calculated as:
$D = \frac{3a-2b}{a+b} - \frac{a-b}{a+b} = \frac{3a-2b-a+b}{a+b} = \frac{2a-b}{a+b}$.
The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}[2A + (n-1)D]$.
For $n = 11$:
$S_{11} = \frac{11}{2} \left[ 2 \left( \frac{a-b}{a+b} \right) + (11-1) \left( \frac{2a-b}{a+b} \right) \right]$
$S_{11} = \frac{11}{2} \left[ \frac{2a-2b}{a+b} + \frac{10(2a-b)}{a+b} \right]$
$S_{11} = \frac{11}{2} \left[ \frac{2a-2b + 20a - 10b}{a+b} \right]$
$S_{11} = \frac{11}{2} \left[ \frac{22a - 12b}{a+b} \right]$
$S_{11} = \frac{11 \cdot 2(11a - 6b)}{2(a+b)} = \frac{11(11a - 6b)}{a+b} = \frac{121a - 66b}{a+b}$.