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 $a_1$.

MEMEG0D Aug 25, 2024

#1**0 **

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**

Maxematics Aug 26, 2024