+118 Let a_1, a_2, a_3, ..., a_11, a_12 be an arithmetic sequence. If \(a_1 + a_3 + a_5 + a_7 + a_9 + a_{11} = -2\) and \(a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 1\), then find a_1.
+1911 Let's denote the common difference of the arithmetic sequence by \(d\).
For the sum of an arithmetic series, we have the formula:
\[ S = \frac{n}{2} \cdot (a_1 + a_n) \]
Given that \(a_1 + a_3 + a_5 + a_7 + a_9 + a_{11} = -2\), we can write the sum as:
\[ S_1 = \frac{6}{2} \cdot (a_1 + a_{11}) = 3 \cdot (a_1 + (a_1 + 10d)) = 3 \cdot (2a_1 + 10d) \]
\[ S_1 = 6a_1 + 30d \]
Similarly, for \(a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 1\), we get:
\[ S_2 = \frac{6}{2} \cdot (a_2 + a_{12}) = 3 \cdot (a_2 + (a_2 + 11d)) = 3 \cdot (2a_2 + 11d) \]
\[ S_2 = 6a_2 + 33d \]
Given that \(a_1 + a_3 + a_5 + a_7 + a_9 + a_{11} = -2\) and \(a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 1\), we can write the following system of equations:
\[ 6a_1 + 30d = -2 \]
\[ 6a_2 + 33d = 1 \]
Now, let's solve this system of equations. Subtracting the first equation from the second, we get:
\[ 6a_2 + 33d - (6a_1 + 30d) = 1 - (-2) \]
\[ 6a_2 - 6a_1 + 3d = 3 \]
\[ 6(a_2 - a_1) + 3d = 3 \]
\[ 6d + 3d = 3 \]
\[ 9d = 3 \]
\[ d = \frac{1}{3} \]
Now, substitute \(d = \frac{1}{3}\) into one of the original equations to find \(a_1\).
Let's use the first equation:
\[ 6a_1 + 30 \left(\frac{1}{3}\right) = -2 \]
\[ 6a_1 + 10 = -2 \]
\[ 6a_1 = -12 \]
\[ a_1 = \boxed{-2} \]
So, \(a_1 = -2\).
+118 I got it!
Let x represent the common difference.
\(a_1+a_3+a_5+a_7+a_9+a_{11} = \frac{6(a_1+a_1+10x)}{2} = 6a_1+30x = -2 \)
\(a_2+a_4+a_6+a_8+a_{10}+a_{12} = \frac{6(a_1+x+a_1+11x)}{2} = 6a_1+36x = 1\)
Solving the equation gives us \(x = \frac{1}{2}, a_1 = -\frac{17}{6}\)
