Algebra

If $d$ is the common difference between consecutive terms of the arithmetic sequence, then we can rewrite the sequaence as $a_1, a_1 + d, a_1 + 2d, \dots, a_1+7d, a_1 + 8d, a_1 + 9d$.

We can now werite both equations as

 $1: a_1 + (a_1 + 2d) = 5$ and $2: (a_1 + d) + (a_1 + 3d) = 6$ and solve for d and a.

Let's use elimination:

$2: 2a_1 + 4d = 6$

$1: 2a_1 + 2d = 5$

$2d = 1$

$d = \frac{1}{2}$

$1: 2a_1 + 2(\frac{1}{2}) = 5$

$2a_1 + 1 = 5$

$2a_1 = 4$

$a_1 = 2$

This means that the sequence is 2, 2.5, 3, 3.5, ... and that a_1 = 2