一个等差数列的首项为 $a$，公差为 $d$，其中 $a$ 和 $d$ 均非零。如果该等差数列的第一项、第五项和第十一项分别等于一个等比数列的第一项、第二项和第三项，求该等比数列前十项的和的精确值，并表示为 $ka$ 的形式，其中 $k$ 是一个最简有理数。

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 $S_{10} = \frac{a(r^{10} - 1)}{r - 1}$:
$S_{10} = \frac{a\left(\left(\frac{3}{2}\right)^{10} - 1\right)}{\frac{3}{2} - 1}$
$S_{10} = \frac{a\left(\frac{59049}{1024} - 1\right)}{\frac{1}{2}}$
$S_{10} = 2a\left(\frac{59049 - 1024}{1024}\right)$
$S_{10} = 2a\left(\frac{58025}{1024}\right)$
$S_{10} = \frac{58025}{512}a$