\( \sum _ { r = 1 } ^ { m } 5 ^ { r } = 19530 \)

To solve the equation: \[ \sum_{r=1}^{m} 5^{r} = 19,\!530 \] we can use the formula for the sum of a geometric series. The sum of the first \( m \) terms of a geometric series with the first term \( a \) and common ratio \( r \) is given by: \[ S = a \frac{r^{m} - 1}{r - 1} \] In this case: - The first term \( a = 5 \) - The common ratio \( r = 5 \) - The sum \( S = 19,\!530 \) Plugging these values into the formula: \[ 19,\!530 = 5 \frac{5^{m} - 1}{5 - 1} \] Simplifying the equation: \[ 19,\!530 = 5 \frac{5^{m} - 1}{4} \] \[ 19,\!530 \times 4 = 5 (5^{m} - 1) \] \[ 78,\!120 = 5 (5^{m} - 1) \] \[ \frac{78,\!120}{5} = 5^{m} - 1 \] \[ 15,\!624 = 5^{m} - 1 \] \[ 5^{m} = 15,\!625 \] Recognizing that \( 15,\!625 \) is a power of 5: \[ 5^{6} = 15,\!625 \] Therefore: \[ m = 6 \] **Answer:** \( m = 6 \)