\( p: 2 p+2: 5 p+3: \ldots \) are the first three terms of an arithmetic sequen (a) Calculate the value of \( p \). (b) Determine the sequence. (c) Find the 49 th term. (d) Which term of the sequence is \( 100 \frac{1}{2} \) ?

To solve the given arithmetic sequence, we start with the terms: 1. The first term is \( p \). 2. The second term is \( 2p + 2 \). 3. The third term is \( 5p + 3 \). ### (a) Calculate the value of \( p \). In an arithmetic sequence, the difference between consecutive terms is constant. Therefore, the difference between the second term and the first term should equal the difference between the third term and the second term: \[ (2p + 2) - p = (5p + 3) - (2p + 2) \] Calculating both sides gives: \[ p + 2 = 3p + 1 \] Now, rearranging this equation: \[ p + 2 - 1 = 3p \] \[ 1 = 3p - p \] \[ 1 = 2p \] \[ p = \frac{1}{2} \] ### (b) Determine the sequence. Using \( p = \frac{1}{2} \): 1. First term: \( p = \frac{1}{2} \) 2. Second term: \( 2p + 2 = 2(\frac{1}{2}) + 2 = 1 + 2 = 3 \) 3. Third term: \( 5p + 3 = 5(\frac{1}{2}) + 3 = \frac{5}{2} + 3 = \frac{5}{2} + \frac{6}{2} = \frac{11}{2} \) Now calculating the common difference: \[ 3 - \frac{1}{2} = \frac{6}{2} - \frac{1}{2} = \frac{5}{2} \] \[ \frac{11}{2} - 3 = \frac{11}{2} - \frac{6}{2} = \frac{5}{2} \] Thus, the sequence is: \( \frac{1}{2}, 3, \frac{11}{2}, \ldots \) ### (c) Find the 49th term. The \( n \)-th term of an arithmetic sequence can be calculated using the formula: \[ a_n = a_1 + (n - 1)d \] Where \( a_1 = \frac{1}{2} \) and \( d = \frac{5}{2} \): \[ a_{49} = \frac{1}{2} + (49 - 1)\frac{5}{2} \] \[ = \frac{1}{2} + 48 \cdot \frac{5}{2} \] \[ = \frac{1}{2} + \frac{240}{2} \] \[ = \frac{241}{2} \] ### (d) Which term of the sequence is \( 100 \frac{1}{2} \)? First, convert \( 100 \frac{1}{2} \) to an improper fraction: \[ 100 \frac{1}{2} = 100.5 = \frac{201}{2} \] We set this equal to the \( n \)-th term formula: \[ \frac{201}{2} = \frac{1}{2} + (n - 1) \frac{5}{2} \] Isolate \( n \): \[ \frac{201}{2} - \frac{1}{2} = (n - 1) \frac{5}{2} \] \[ \frac{200}{2} = (n - 1) \frac{5}{2} \] \[ 100 = (n - 1) \cdot \frac{5}{2} \] Multiplying through by \( 2 \): \[ 200 = 5(n - 1) \] \[ 40 = n - 1 \] \[ n = 41 \] Thus, \( 100 \frac{1}{2} \) is the \( 41^{st} \) term of the sequence.