Calculate the value of \( p \) if \( \sum_{k=0}^{\infty} 27 p^{k+1}=\sum_{r=3}^{14}(30-3 r) \)

To solve the equation \[ \sum_{k=0}^{\infty} 27 p^{k+1} = \sum_{r=3}^{14}(30-3 r), \] we start by evaluating the right side, \( \sum_{r=3}^{14} (30 - 3r) \). 1. First, we find the expression for \( 30 - 3r \) as \( r \) varies from 3 to 14: - When \( r = 3 \): \( 30 - 3(3) = 30 - 9 = 21 \) - When \( r = 4 \): \( 30 - 3(4) = 30 - 12 = 18 \) - When \( r = 5 \): \( 30 - 3(5) = 30 - 15 = 15 \) - ... - When \( r = 14 \): \( 30 - 3(14) = 30 - 42 = -12 \) 2. The terms created give the sequence: \[ 21, 18, 15, 12, 9, 6, 3, 0, -3, -6, -9, -12. \] 3. We calculate the sum directly: \[ \sum_{r=3}^{14}(30-3 r) = 21 + 18 + 15 + 12 + 9 + 6 + 3 + 0 - 3 - 6 - 9 - 12. \] Grouping these terms helps: - Positive Terms: \( 21 + 18 + 15 + 12 + 9 + 6 + 3 = 84 \) - Negative Terms: \( 0 - 3 - 6 - 9 - 12 = -30 \) 4. Now we complete the sum: \[ 84 - 30 = 54. \] Thus, we have: \[ \sum_{r=3}^{14}(30-3 r) = 54. \] Next, we turn our attention to the left side, \( \sum_{k=0}^{\infty} 27 p^{k+1} \): This can be rewritten as: \[ 27 p^{1} \sum_{k=0}^{\infty} p^{k} = 27p \cdot \frac{1}{1 - p} \quad \text{(for } |p| < 1\text{)} \] Setting this equal to our previously computed sum: \[ 27p \cdot \frac{1}{1 - p} = 54. \] Now we simplify and solve for \( p \): 1. Cross-multiply to eliminate the fraction: \[ 27p = 54(1 - p). \] 2. Distribute on the right-hand side: \[ 27p = 54 - 54p. \] 3. Add \( 54p \) to both sides: \[ 27p + 54p = 54, \] which simplifies to: \[ 81p = 54. \] 4. Finally, divide both sides by 81: \[ p = \frac{54}{81} = \frac{2}{3}. \] Thus, the value of \( p \) is \[ \boxed{\frac{2}{3}}. \]