YEARBOOKS The yearbook staff is unpacking a box of school yearbooks. The arithmetic sequence \( 281,270,259,248 \ldots \) represents the total number of ounces that the box weighs as each yearbook is taken out of the box. a. Write a function to represent this sequence. b. Determine the weight of each yearbook. c. If the box weighs at least 11 ounces empty and 292 ounces when it is full, how many yearbooks were in the box?

To solve the problem step by step, let's extract the known conditions from the arithmetic sequence provided. ### Known Conditions: 1. The sequence of weights is \( 281, 270, 259, 248, \ldots \). 2. This is an arithmetic sequence where the first term \( a_1 = 281 \) and the common difference \( d = 270 - 281 = -11 \). 3. The box weighs at least 11 ounces empty. 4. The box weighs 292 ounces when it is full. ### Step a: Write a function to represent this sequence. The general formula for the \( n \)-th term of an arithmetic sequence can be expressed as: \[ a_n = a_1 + (n - 1) \cdot d \] Substituting the known values: - \( a_1 = 281 \) - \( d = -11 \) The function representing the sequence is: \[ a_n = 281 + (n - 1)(-11) \] ### Step b: Determine the weight of each yearbook. To find the weight of each yearbook, we need to determine the difference between consecutive terms in the sequence, which we already identified as the common difference \( d = -11 \). Thus, the weight of each yearbook is: \[ \text{Weight of each yearbook} = 11 \text{ ounces} \] ### Step c: If the box weighs at least 11 ounces empty and 292 ounces when it is full, how many yearbooks were in the box? To find the number of yearbooks, we can set up the equation based on the total weight of the box when full and empty. Let \( n \) be the number of yearbooks. The total weight of the box when full is given by: \[ \text{Weight of box when full} = \text{Weight of empty box} + n \cdot \text{Weight of each yearbook} \] Given: - Weight of empty box \( \geq 11 \) ounces - Weight of box when full = 292 ounces We can express this as: \[ 292 = \text{Weight of empty box} + n \cdot 11 \] Since the weight of the empty box is at least 11 ounces, we can set the weight of the empty box to 11 ounces for the maximum number of yearbooks: \[ 292 = 11 + n \cdot 11 \] \[ 292 - 11 = n \cdot 11 \] \[ 281 = n \cdot 11 \] \[ n = \frac{281}{11} \] Now, let's calculate \( n \). Calculate the value by following steps: - step0: Calculate: \(\frac{281}{11}\) Calculating \( n \): \[ n = \frac{281}{11} \approx 25.5454 \] Since \( n \) must be a whole number (as you cannot have a fraction of a yearbook), we round down to the nearest whole number: \[ n = 25 \] ### Conclusion: - **a.** The function representing the sequence is \( a_n = 281 + (n - 1)(-11) \). - **b.** The weight of each yearbook is \( 11 \) ounces. - **c.** The number of yearbooks in the box is \( 25 \).