The given series is $\frac{1}{16807}, \frac{1} {2401}, \frac{1}{343}, \frac{1}{49}, ...$ This is a geometric series. Let the first term be $a_1$ and the common ratio be $r$. $a_1 = \frac{1}{16807}$. The common ratio $r$ is found by dividing any term by its preceding term. $r = \frac{1/2401} {1/16807} = \frac{16807}{2401}$. We recognize that the denominators are powers of 7: $49 = 7^2$ $343 = 7^3$ $2401 = 7^4$ $16807 = 7^5$ So the series is $\frac{1}{7^5}, \frac{1}{7^4}, \frac{1} {7^3}, \frac{1}{7^2}, ...$ The first term is $a_1 = \frac{1}{7^5}$. The common ratio is $r = \frac{1/7^4}{1/7^5} = \frac{7^5}{7^4} = 7$. The formula for the $n$-th term of a geometric series is $a_n = a_1 \cdot r^{n-1}$. We need to find the 10th term, so $n=10$. $a_{10} = a_1 \cdot r^{10-1} = a_1 \cdot r^9$. $a_{10} = \frac{1}{7^5} \cdot 7^9 = 7^{9-5} = 7^4$. Calculate $7^4$: $7^1 = 7$ $7^2 = 49$ $7^3 = 343$ $7^4 = 2401$. The 10th term is 2401.