25. Determine \( r \) if \( \sum_{n=1}^{\infty} 2 r^{n-1}=12 \). (5)

To determine the value of \( r \) that satisfies the equation: \[ \sum_{n=1}^{\infty} 2 r^{n-1} = 12 \] we recognize this as an infinite geometric series. **Step 1: Identify the Geometric Series Parameters** A geometric series has the form: \[ \sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r} \quad \text{for} \ |r| < 1 \] In the given problem, the series starts at \( n = 1 \), so we can adjust the formula accordingly: \[ \sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1 - r} \] Comparing with the given series: \[ \sum_{n=1}^{\infty} 2 r^{n-1} \quad \Rightarrow \quad a = 2 \] **Step 2: Apply the Geometric Series Formula** Using the formula: \[ \frac{a}{1 - r} = 12 \] Substitute \( a = 2 \): \[ \frac{2}{1 - r} = 12 \] **Step 3: Solve for \( r \)** Multiply both sides by \( 1 - r \): \[ 2 = 12(1 - r) \] Divide both sides by 12: \[ \frac{2}{12} = 1 - r \quad \Rightarrow \quad \frac{1}{6} = 1 - r \] Subtract \( \frac{1}{6} \) from both sides: \[ r = 1 - \frac{1}{6} = \frac{5}{6} \] **Final Answer:** \( r = \dfrac{5}{6} \)