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# Arithmetic Sequence

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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.

Mar 30, 2024

#1
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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, $$a_1 = -2$$.

Mar 30, 2024
#2
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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 $$x = \frac{1}{2}, a_1 = -\frac{17}{6}$$

Mar 30, 2024