We are given a geometric sequence ${a_n}$ with the sum of the first $n$ terms denoted as $S_n$. We know that $a_1 + a_3 = 5$ and $S_4 = 15$. We need to find $S_6$.

1. Problem Description

We are given a geometric sequence ${a_n}$ with the sum of the first $n$ terms denoted as $S_n$. We know that $a_1 + a_3 = 5$ and $S_4 = 15$. We need to find $S_6$.

2. Solution Steps

Let $a_1 = a$ and the common ratio be $q$.

Then $a_n = aq^{n-1}$.

We are given $a_1 + a_3 = 5$, which can be written as

$a + aq^2 = 5$

$a(1+q^2) = 5$ (1)

We are also given $S_4 = 15$, which can be written as

$S_4 = \frac{a(1-q^4)}{1-q} = 15$

$a(1-q^4) = 15(1-q)$ (2)

$a(1-q^2)(1+q^2) = 15(1-q)$

From equation (1), we have $a(1+q^2) = 5$. Substitute this into the above equation:

$5(1-q^2) = 15(1-q)$

$1-q^2 = 3(1-q)$

$(1-q)(1+q) = 3(1-q)$

If $q = 1$, then $a + a = 5$, so $a = \frac{5}{2}$. Also $S_4 = 4a = 15$, so $a = \frac{15}{4}$. This is a contradiction. Thus $q \ne 1$.

Therefore, $1+q = 3$, so $q = 2$.

Substitute $q = 2$ into equation (1):

$a(1+2^2) = 5$

$5a = 5$, so $a = 1$.

We want to find $S_6$.

$S_6 = \frac{a(1-q^6)}{1-q} = \frac{1(1-2^6)}{1-2} = \frac{1-64}{-1} = \frac{-63}{-1} = 63$

3. Final Answer

The final answer is D. 63