1. **Problem statement:** Given the sum of a geometric series $ \sum_{p=k}^{10} 3^{p-1} = 29520 $, find the value of $ k $. 2. **Formula for the sum of a geometric series:** $$ S_n = a \frac{r^n - 1}{r - 1} $$ where $ a $ is the first term, $ r $ is the common ratio, and $ n $ is the number of terms. 3. **Identify terms:** - The series is $ 3^{k-1} + 3^k + 3^{k+1} + \cdots + 3^9 $ (since upper limit is 10, last term is $ p=10 $ so exponent is $ 10-1=9 $). - Number of terms $ n = 10 - k + 1 = 11 - k $. - First term $ a = 3^{k-1} $. - Common ratio $ r = 3 $. 4. **Write the sum using the formula:** $$ 29520 = 3^{k-1} \cdot \frac{3^{11-k} - 1}{3 - 1} = 3^{k-1} \cdot \frac{3^{11-k} - 1}{2} $$ 5. **Multiply both sides by 2:** $$ 2 \times 29520 = 3^{k-1} (3^{11-k} - 1) $$ $$ 59040 = 3^{k-1} (3^{11-k} - 1) $$ 6. **Simplify the right side:** $$ 3^{k-1} \cdot 3^{11-k} = 3^{(k-1)+(11-k)} = 3^{10} $$ So, $$ 59040 = 3^{10} - 3^{k-1} $$ 7. **Rearrange to isolate $ 3^{k-1} $:** $$ 3^{k-1} = 3^{10} - 59040 $$ 8. **Calculate $ 3^{10} $:** $$ 3^{10} = 59049 $$ 9. **Substitute:** $$ 3^{k-1} = 59049 - 59040 = 9 $$ 10. **Solve for $ k $:** Since $ 3^{k-1} = 9 = 3^2 $, $$ k - 1 = 2 \implies k = 3 $$ **Final answer:** $ k = 3 $