+238 Let a_1, a_2, a_3, \dots, a_8, a_9, a_{10} be an arithmetic sequence. If $a_1 + a_3 = 5$ and $a_2 + a_4 = 6$, then find $a_1$.
+135 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
