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 $a_1$.

bader Feb 18, 2024

#1**+1 **

Rewrite this arithmetic sequence as \(a_{1}, a_{1}+r, a_{1}+2r, \dots, a_{1}+10r, a_{1}+11r\), where a_{1} 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.

hairyberry Feb 18, 2024