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

To solve for \( p \) in the given equation \( \sum_{k=0}^{\infty} 27 p^{k+1} = \sum_{r=3}^{14}(30-3r) \), we will start by simplifying each side of the equation. **Step 1: Simplifying the Left Side** The left side, \( \sum_{k=0}^{\infty} 27 p^{k+1} \), can be factored out as follows: \[ \sum_{k=0}^{\infty} 27 p^{k+1} = 27 p \sum_{k=0}^{\infty} p^k \] The series \( \sum_{k=0}^{\infty} p^k \) is a geometric series that converges if \( |p| < 1 \): \[ \sum_{k=0}^{\infty} p^k = \frac{1}{1-p} \] Thus, the left side becomes: \[ 27 p \cdot \frac{1}{1-p} = \frac{27p}{1-p} \] **Step 2: Simplifying the Right Side** Now we simplify the right side, \( \sum_{r=3}^{14}(30-3r) \). We can rewrite this as: \[ \sum_{r=3}^{14}(30-3r) = \sum_{r=3}^{14}30 - \sum_{r=3}^{14}3r \] Calculating \( \sum_{r=3}^{14}30 \): - There are \( 14 - 3 + 1 = 12 \) terms. \[ \sum_{r=3}^{14}30 = 30 \times 12 = 360 \] Now calculating \( \sum_{r=3}^{14}3r \): \[ \sum_{r=3}^{14}3r = 3 \sum_{r=3}^{14} r \] The sum \( \sum_{r=3}^{14} r \) is a series of integers from 3 to 14, which can be calculated as: \[ \sum_{r=3}^{14} r = \frac{n}{2}(a + l) \] where \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term. - \( n = 14 - 3 + 1 = 12 \) - \( a = 3 \), \( l = 14 \) Calculating: \[ \sum_{r=3}^{14} r = \frac{12}{2}(3 + 14) = 6 \times 17 = 102 \] Thus, \[ \sum_{r=3}^{14}3r = 3 \times 102 = 306 \] Putting it all together on the right side, we have: \[ \sum_{r=3}^{14}(30-3r) = 360 - 306 = 54 \] **Step 3: Setting Both Sides Equal** Now, we set both expressions equal to each other: \[ \frac{27p}{1-p} = 54 \] **Step 4: Solving for \( p \)** Cross-multiplying gives: \[ 27p = 54(1-p) \] Simplifying further: \[ 27p = 54 - 54p \] Combining like terms: \[ 27p + 54p = 54 \] This simplifies to: \[ 81p = 54 \] Dividing both sides by 81 results in: \[ p = \frac{54}{81} = \frac{2}{3} \] Thus, the value of \( p \) is: \[ \boxed{\frac{2}{3}} \]