In an arithmetic sequence $\{a_n\}$, if $a_1 + a_3 + a_5 = 105$ and $a_2 + a_4 + a_6 = 99$, find $a_{20}$.

1. Problem Description

In an arithmetic sequence $\{a_n\}$, if $a_1 + a_3 + a_5 = 105$ and $a_2 + a_4 + a_6 = 99$, find $a_{20}$.

2. Solution Steps

Let $d$ be the common difference of the arithmetic sequence $\{a_n\}$. Then $a_n = a_1 + (n-1)d$.

We are given:

$a_1 + a_3 + a_5 = 105$

$a_2 + a_4 + a_6 = 99$

Using the formula for the nth term, we have:

$a_1 + (a_1 + 2d) + (a_1 + 4d) = 105$

$3a_1 + 6d = 105$

$a_1 + 2d = 35$ (Dividing by 3)

Also,

$a_2 + a_4 + a_6 = 99$

$(a_1 + d) + (a_1 + 3d) + (a_1 + 5d) = 99$

$3a_1 + 9d = 99$

$a_1 + 3d = 33$ (Dividing by 3)

Now we have two equations:

$a_1 + 2d = 35$

$a_1 + 3d = 33$

Subtracting the first equation from the second, we get:

$(a_1 + 3d) - (a_1 + 2d) = 33 - 35$

$d = -2$

Substituting $d = -2$ into the first equation:

$a_1 + 2(-2) = 35$

$a_1 - 4 = 35$

$a_1 = 39$

Now we can find $a_{20}$ using the formula $a_n = a_1 + (n-1)d$:

$a_{20} = a_1 + (20-1)d$

$a_{20} = 39 + 19(-2)$

$a_{20} = 39 - 38$

$a_{20} = 1$

3. Final Answer

$a_{20} = 1$