tics Mathematics Grade 12 Learner Material What is the value of \( m \) for which: \[ \sum_{k=1}^{m} 5(3)^{k-1}=65 \] Consider the sequence:

To determine the value of \( m \) that satisfies the equation: \[ \sum_{k=1}^{m} 5 \cdot 3^{k-1} = 65 \] we can recognize that this sum represents a **geometric series**. Let's break down the solution step-by-step. ### Step 1: Identify the Components of the Geometric Series A geometric series has the general form: \[ S_m = a \cdot \frac{r^m - 1}{r - 1} \] where: - \( S_m \) is the sum of the first \( m \) terms, - \( a \) is the first term, - \( r \) is the common ratio, - \( m \) is the number of terms. For the given series: \[ \sum_{k=1}^{m} 5 \cdot 3^{k-1} \] - **First Term (\( a \))**: When \( k = 1 \), \[ a = 5 \cdot 3^{1-1} = 5 \cdot 3^{0} = 5 \cdot 1 = 5 \] - **Common Ratio (\( r \))**: The ratio between successive terms is \( 3 \) since each term is multiplied by \( 3 \) to get the next term. ### Step 2: Plug the Values into the Sum Formula Using the formula for the sum of a geometric series: \[ S_m = a \cdot \frac{r^m - 1}{r - 1} \] Substitute the known values: \[ 65 = 5 \cdot \frac{3^m - 1}{3 - 1} \] \[ 65 = 5 \cdot \frac{3^m - 1}{2} \] ### Step 3: Solve for \( m \) 1. **Multiply both sides by 2 to eliminate the denominator:** \[ 65 \times 2 = 5 \cdot (3^m - 1) \] \[ 130 = 5 \cdot (3^m - 1) \] 2. **Divide both sides by 5:** \[ \frac{130}{5} = 3^m - 1 \] \[ 26 = 3^m - 1 \] 3. **Add 1 to both sides:** \[ 26 + 1 = 3^m \] \[ 27 = 3^m \] 4. **Find the value of \( m \) such that \( 3^m = 27 \):** Since \( 3^3 = 27 \), \[ m = 3 \] ### Conclusion The value of \( m \) that satisfies the equation is: \[ \boxed{3} \]