+327 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 $a_1$.
+1926 Let's do a very sneaky trick to solve this problem. Now, if we add the two sequences together, we get
\(a_1 + a_3 + a_5 + a_7 + a_9 + a_{11} = 0 \\ a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 0 \space + \)
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\(a_1+ a_2+ a_3+ \dots+ a_{10}+ a_{11}+ a_{12}=0\)
Now, this sum is really important.
Now, first, let's consider the possibility of something where half the numbers are negative, and the other half is the absolute value of the numbers.
Although this is possible, we have an even number of terms, meaning it's impossible to account for 0.
EX: \(-5, -4, -3,-2,-1,0,1,2,3,4,5 (6 \text{ extra})\)
Thus, the only possibility is EVERY SINGLE TERM is 0.
This satisfies everything, so meaning that \(a_1=0\)
So 0 is our answer.
I hope I'm right...
Thanks! :)
