Siven arithmetic series \( 18+24+30+\cdots+300 \) (a) Determine the number of terms in the series. (b) Calculate the sum of this series. (c) Calculate the sum of all the whole numbers up to and including 300 that GAUTENG PROVINCE tixamenss © Gauteng Department of Education Page 6 of are not divisible by six. (4) QUESTION 9

To tackle the arithmetic series \( 18 + 24 + 30 + \cdots + 300 \), we first observe that this series has a common difference of \( 6 \). The first term \( a = 18 \) and the last term \( l = 300 \). (a) To find the number of terms \( n \), we use the formula for the \( n \)-th term of an arithmetic series: \[ l = a + (n - 1)d \] Substituting the known values: \[ 300 = 18 + (n - 1) \cdot 6 \] Solving for \( n \): \[ 300 - 18 = (n - 1) \cdot 6 \\ 282 = (n - 1) \cdot 6 \\ n - 1 = 47 \\ n = 48 \] Thus, there are \( 48 \) terms in the series. (b) To calculate the sum \( S_n \) of the series, we use the formula: \[ S_n = \frac{n}{2} (a + l) \] Substituting the values we've found: \[ S_{48} = \frac{48}{2} (18 + 300) \\ = 24 \cdot 318 \\ = 7632 \] Therefore, the sum of the series is \( 7632 \). (c) Now to calculate the sum of all whole numbers up to \( 300 \) that are not divisible by \( 6 \). First, we find the sum of numbers divisible by \( 6 \): The numbers divisible by \( 6 \) up to \( 300 \) form another arithmetic series: \( 6, 12, 18, \ldots, 300 \) with a common difference of \( 6 \) too. The last term \( 300 = 6n \) implies \( n = 50 \). Therefore, there are \( 50 \) terms. The sum of this series is: \[ S_{50} = \frac{50}{2} (6 + 300) = 25 \cdot 306 = 7650 \] Next, the total sum of numbers from \( 1 \) to \( 300 \) is: \[ \text{Total Sum} = \frac{300}{2}(1 + 300) = 150 \cdot 301 = 45150 \] To find the sum of numbers not divisible by \( 6 \): \[ \text{Sum Not Divisible by 6} = \text{Total Sum} - S_{50} = 45150 - 7650 = 37500 \] Thus, the sum of all whole numbers up to and including \( 300 \) that are not divisible by \( 6 \) is \( 37500 \).