The problem states that the twentieth term of the sequence $m, \frac{2}{3}m, \frac{1}{3}m, \dots$ is 15. We need to find the value of $m$. This is an arithmetic sequence.

1. Problem Description

The problem states that the twentieth term of the sequence $m, \frac{2}{3}m, \frac{1}{3}m, \dots$ is

1

5. We need to find the value of $m$. This is an arithmetic sequence.

2. Solution Steps

The sequence is given by $m, \frac{2}{3}m, \frac{1}{3}m, \dots$.

The first term is $a_1 = m$.

The common difference $d$ can be found by subtracting the first term from the second term:

$d = \frac{2}{3}m - m = \frac{2}{3}m - \frac{3}{3}m = -\frac{1}{3}m$.

The formula for the $n$-th term of an arithmetic sequence is:

$a_n = a_1 + (n-1)d$.

In this case, we are given that the 20th term is 15, so $a_{20} = 15$ and $n = 20$. Substituting the known values into the formula, we get:

$15 = m + (20 - 1)\left(-\frac{1}{3}m\right)$

$15 = m + 19\left(-\frac{1}{3}m\right)$

$15 = m - \frac{19}{3}m$

$15 = \frac{3}{3}m - \frac{19}{3}m$

$15 = -\frac{16}{3}m$

Now we solve for $m$:

$15 = -\frac{16}{3}m$

Multiply both sides by 3:

$45 = -16m$

Divide both sides by -16:

$m = -\frac{45}{16}$

3. Final Answer

$m = -\frac{45}{16}$