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 $a_1$.

tomtom Jun 15, 2024

#1**+1 **

We can use a clever trick to solve this problem.

We have two equations. Subtracting the first equation from the second equation, we get

\((a_2 - a_1) + (a_4 - a_3) + (a_6 - a_5) + (a_8 - a_7) + (a_{10} - a_9) + (a_{12} - a_{11}) = 0\)

The reason why this is important is because in every parenthesis, it forms the common difference d.

Thus, we have

\(6d=0\\ d=0\)

This means every number in the sequence is the same.

Since the entire sequence is equal to 0, we have \(a_1 = 0\)

So our answer is 0.

Thanks! :)

NotThatSmart Jun 15, 2024

#1**+1 **

Best Answer

We can use a clever trick to solve this problem.

We have two equations. Subtracting the first equation from the second equation, we get

\((a_2 - a_1) + (a_4 - a_3) + (a_6 - a_5) + (a_8 - a_7) + (a_{10} - a_9) + (a_{12} - a_{11}) = 0\)

The reason why this is important is because in every parenthesis, it forms the common difference d.

Thus, we have

\(6d=0\\ d=0\)

This means every number in the sequence is the same.

Since the entire sequence is equal to 0, we have \(a_1 = 0\)

So our answer is 0.

Thanks! :)

NotThatSmart Jun 15, 2024