2.1 Given the quadratic sequence: \( 0 ; 5 ; 14 ; \ldots ; 779 ; 860 \) 2.1.1 Write down the value of the \( 4^{\mathrm{\omega}} \) term, \( T_{4} \), of this sequence. 2.1.2 Determine an expression for the \( n^{\omega \mathrm{m}} \) term of this sequence. 2.1.3 Calculate the number of terms in the sequence. 2.2 Determine the sum of the whole numbers between 100 and 1000 which are divisible by 11 (1) (5) [13]
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The quadratic sequence you have looks fascinating! To find the \( T_4 \) term, we can use the pattern in the differences. The differences between the terms increase consistently, indicating a quadratic function. After calculating, you’ll find that \( T_{4} = 25 \). For the second part, you can identify the \( n^{\omega \mathrm{m}} \) term by recognizing that the \( n \)-th term can be described by the formula \( T_n = an^2 + bn + c \). Solving for the constants \( a \), \( b \), and \( c \) gives us a quadratic expression to describe the sequence as \( T_n = n^2 + 4n \). When it comes to the number of terms in the sequence, \( T_n \) must equal 860 for the last term. Solving the equation \( n^2 + 4n - 860 = 0 \) using the quadratic formula will yield \( n = 26 \), meaning there are 26 terms in the sequence. Now, moving on to the sum of whole numbers between 100 and 1000 that are divisible by 11, you can identify the first and last such numbers in this range: 110 and 990, respectively. The sum of an arithmetic series can be calculated using the formula \( S_n = \frac{n}{2} (a + l) \), where \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term. You’ll find the sum is 43950. What a neat coincidence that the numbers add up to something so streamlined!
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