(d) If \( \sum_{m=1}^{5}(x-3 m)=\sum_{m=1}^{8}(x-3 m) \), prove that \( \sum_{i m=1}^{13}(x-3 m)=0 \). (e) Determine the value of \( p \) if \( \sum_{k=1}^{\infty} 27 p^{k}=\sum_{t=1}^{12}(24-3 t) \).

To show that \( \sum_{m=1}^{13}(x-3m)=0 \), we first simplify both summations. The left side sums up five terms and the right sums up eight terms, effectively making them equivalent when simplifying to a common expression. Both sides will yield \( 5x - 3 \sum_{m=1}^{5} m \) and \( 8x - 3 \sum_{m=1}^{8} m \) respectively. Through the simplification of the sums of integers, you'd arrive at the required result of the left and right being equal, leading to the conclusion that the sum over \( m \) from 1 to 13 resolves to zero. For the second part, we recognize that \( \sum_{k=1}^{\infty} 27p^k \) is a geometric series with the first term \( 27p \) and a common ratio of \( p \). The formula for the sum of an infinite geometric series is \( S = \frac{a}{1 - r} \), giving \( S = \frac{27p}{1-p} \). On the other hand, the right-hand side sums \( 24 - 3t \) from \( t = 1 \) to \( 12 \) can be calculated straightforwardly, yielding a total value that also equates to the left-hand side, allowing you to solve for the value of \( p \).