In an arithmetic sequence $\{a_n\}$, if $3(a_3 + a_5) + 2(a_7 + a_{10} + a_{13}) = 24$, find $S_{13}$.

1. Problem Description

In an arithmetic sequence $\{a_n\}$, if $3(a_3 + a_5) + 2(a_7 + a_{10} + a_{13}) = 24$, find $S_{13}$.

2. Solution Steps

Since $\{a_n\}$ is an arithmetic sequence, we can express each term using the first term $a_1$ and the common difference $d$. Thus, $a_n = a_1 + (n-1)d$.

Then

$a_3 = a_1 + 2d$

$a_5 = a_1 + 4d$

$a_7 = a_1 + 6d$

$a_{10} = a_1 + 9d$

$a_{13} = a_1 + 12d$

Substituting these into the given equation:

$3(a_1 + 2d + a_1 + 4d) + 2(a_1 + 6d + a_1 + 9d + a_1 + 12d) = 24$

$3(2a_1 + 6d) + 2(3a_1 + 27d) = 24$

$6a_1 + 18d + 6a_1 + 54d = 24$

$12a_1 + 72d = 24$

$a_1 + 6d = 2$

Now we need to find $S_{13}$. The sum of an arithmetic series is given by:

$S_n = \frac{n}{2}(a_1 + a_n)$

$S_{13} = \frac{13}{2}(a_1 + a_{13})$

$S_{13} = \frac{13}{2}(a_1 + a_1 + 12d)$

$S_{13} = \frac{13}{2}(2a_1 + 12d)$

$S_{13} = 13(a_1 + 6d)$

Since $a_1 + 6d = 2$, we have:

$S_{13} = 13(2)$

$S_{13} = 26$

3. Final Answer

$S_{13} = 26$