An arithmetic series has first term $a$ and common difference $d$, where $a$ and $d$ are non-zero. If the first, fifth and eleventh terms of the arithmetic series are equal to the first, second and third terms of a geometric series respectively, find the exact sum of the first ten terms of the geometric series, in the form $ka$, where $k$ is a simplified rational number.

1. Set up an equation by expressing the terms using the arithmetic progression and equating the ratio of consecutive terms for the geometric progression.

Let the terms of the geometric series be $T_1$, $T_2$, and $T_3$. From the arithmetic progression, we have:
$T_1 = a$
$T_2 = a + 4d$
$T_3 = a + 10d$

Since these terms form a geometric series, the ratio between consecutive terms must be equal:
$\frac{a + 4d}{a} = \frac{a + 10d}{a + 4d}$

2. Solve the equation to find a relationship between $d$ and $a$.

Cross-multiplying to solve for $d$:
$(a + 4d)^2 = a(a + 10d)$
$a^2 + 8ad + 16d^2 = a^2 + 10ad$
$16d^2 - 2ad = 0$
$2d(8d - a) = 0$

Since it is given that $d \neq 0$, we can divide by $2d$ (or select the non-zero root):
$8d - a = 0$
$d = \frac{a}{8}$

3. Calculate the common ratio $r$ of the geometric series.

Substitute $d = \frac{a}{8}$ back into the ratio expression:
$r = \frac{a + 4\left(\frac{a}{8}\right)}{a}$
$r = \frac{a + \frac{1}{2}a}{a}$
$r = \frac{3}{2}$

4. Use the formula for the sum of a finite geometric series to find the sum of the first ten terms.

The sum of the first 10 terms is given by $S_{10} = \frac{a(r^{10} - 1)}{r - 1}$:
$S_{10} = \frac{a\left(\left(\frac{3}{2}\right)^{10} - 1\right)}{\frac{3}{2} - 1}$
$S_{10} = \frac{a\left(\frac{59049}{1024} - 1\right)}{\frac{1}{2}}$
$S_{10} = 2a\left(\frac{59049 - 1024}{1024}\right)$
$S_{10} = 2a\left(\frac{58025}{1024}\right)$
$S_{10} = \frac{58025}{512}a$