+1768 Let $a_1,$ $a_2,$ $a_3,$ $\dots,$ $a_{10},$ $a_{11},$ $a_{12}$ be an arithmetic sequence. If $a_1 + a_3 + a_5 + a_7 + a_9 + a_{11} = 0$ and $a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 0$, then find $a_1$.
+399 Rewrite this arithmetic sequence as \(a_{1}, a_{1}+r, a_{1}+2r, \dots, a_{1}+10r, a_{1}+11r\), where a1 is the first term, and r is the common difference.
\(a_{1} + a_{3} + a_{5} + a_{7} + a_{9} + a_{11}\) is now \(a_{1} + a_{1}+2r + a_{1}+4r + a_{1} + 6r + a_{1} + 8r + a_{1} + 10r\)
\(a_{2} + a_{4} + a_{6} + a_{8} + a_{10} + a_{12}\)is now \(a_{1} + r + a_{1} + 3r \dots+ a_{1} + 11r\)
Each term in the second one has one more r than the first one, so subtracting:
\(6r=0, r=0\).
Sub this back in to equation 1,
\(6a_{1}=0\)
\(a_{1}=0\).
We get this first term is zero. Note, since the common ratio is 0, this sequence is just twelve zeros, but it satisfies the conditions, and is the only sequence to do so. 
