Given an arithmetic sequence ${a_n}$, we have $a_2 + a_4 = 4$ and $a_3 + a_5 = 10$. We need to find the sum of the first 10 terms, $S_{10}$.

1. Problem Description

Given an arithmetic sequence ${a_n}$, we have $a_2 + a_4 = 4$ and $a_3 + a_5 = 10$. We need to find the sum of the first 10 terms, $S_{10}$.

2. Solution Steps

Let $a_1$ be the first term and $d$ be the common difference of the arithmetic sequence.

We can express each term $a_n$ as $a_n = a_1 + (n-1)d$.

From $a_2 + a_4 = 4$, we have $(a_1 + d) + (a_1 + 3d) = 4$, which simplifies to

$2a_1 + 4d = 4$. Dividing by 2, we get

$a_1 + 2d = 2$ (1)

From $a_3 + a_5 = 10$, we have $(a_1 + 2d) + (a_1 + 4d) = 10$, which simplifies to

$2a_1 + 6d = 10$. Dividing by 2, we get

$a_1 + 3d = 5$ (2)

Subtracting equation (1) from equation (2), we get

$(a_1 + 3d) - (a_1 + 2d) = 5 - 2$

$d = 3$

Substituting $d=3$ into equation (1), we get

$a_1 + 2(3) = 2$

$a_1 + 6 = 2$

$a_1 = -4$

The sum of the first $n$ terms of an arithmetic sequence is given by the formula:

$S_n = \frac{n}{2} [2a_1 + (n-1)d]$

We want to find $S_{10}$, so $n=10$.

$S_{10} = \frac{10}{2} [2a_1 + (10-1)d]$

$S_{10} = 5 [2(-4) + 9(3)]$

$S_{10} = 5 [-8 + 27]$

$S_{10} = 5 [19]$

$S_{10} = 95$

3. Final Answer

$S_{10} = 95$