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The following sequence is a combination of an arithmetic sequence and a geometric sequence: \( 5 ; 5 ; 15 ; 10 ; 25 ; 20 ; 35 ; 40 ; \ldots \) \( 4.1 \quad \) Write down the next 2 terms. 4.2 Calculate \( \mathrm{T}_{42}-\mathrm{T}_{41} \). 4.3 Determine the sum of the first 50 terms. 4.4 Prove that ALL the terms of this infinite sequence will be divisible by 5 . (4)

Ask by Conner Christensen. in South Africa
Jan 23,2025

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The next two terms are 45 and 80. ### 4.2 Calculate \( \mathrm{T}_{42} - \mathrm{T}_{41} \) - **\( \mathrm{T}_{41} \):** Since 41 is odd, it follows the arithmetic sequence pattern. \[ \mathrm{T}_{41} = 5 \times 41 = 205 \] - **\( \mathrm{T}_{42} \):** Since 42 is even, it follows the geometric sequence pattern. \[ \mathrm{T}_{42} = 5 \times 2^{(42/2 - 1)} = 5 \times 2^{20} = 5 \times 1,048,576 = 5,242,880 \] **Calculation:** \[ \mathrm{T}_{42} - \mathrm{T}_{41} = 5,242,880 - 205 = 5,242,675 \] **Answer:** \( \mathrm{T}_{42} - \mathrm{T}_{41} = 5,242,675 \). ### 4.3 Determine the Sum of the First 50 Terms To find the sum of the first 50 terms, we'll separate the sequence into its odd and even parts. - **Sum of Odd-Indexed Terms (\( S_{\text{odd}} \)):** - There are 25 odd terms. - The sequence is an arithmetic series with the first term \( a = 5 \) and common difference \( d = 10 \). - **Sum Formula:** \[ S_{\text{odd}} = \frac{n}{2} \times (2a + (n-1)d) = \frac{25}{2} \times (2 \times 5 + 24 \times 10) = 12.5 \times (10 + 240) = 12.5 \times 250 = 3,125 \] - **Sum of Even-Indexed Terms (\( S_{\text{even}} \)):** - There are 25 even terms. - The sequence is a geometric series with the first term \( a = 5 \) and common ratio \( r = 2 \). - **Sum Formula:** \[ S_{\text{even}} = a \times \frac{r^n - 1}{r - 1} = 5 \times \frac{2^{25} - 1}{2 - 1} = 5 \times (33,554,432 - 1) = 5 \times 33,554,431 = 167,772,155 \] **Total Sum:** \[ S_{\text{total}} = S_{\text{odd}} + S_{\text{even}} = 3,125 + 167,772,155 = 167,775,280 \] **Answer:** The sum of the first 50 terms is **167,775,280**. ### 4.4 Prove That ALL the Terms of This Infinite Sequence Will Be Divisible by 5 - **Odd-Indexed Terms:** - Each term is given by \( T_n = 5n \). - Since \( 5n \) is always divisible by 5, all odd-indexed terms are divisible by 5. - **Even-Indexed Terms:** - Each term is given by \( T_n = 5 \times 2^{(n/2 - 1)} \). - \( 5 \times 2^{(n/2 - 1)} \) is always divisible by 5 because it includes the factor 5. **Conclusion:** Every term in the sequence is divisible by 5. **Answer:** All terms of the sequence are divisible by 5.

Solution

Let's analyze the given sequence: \[ 5 ;\ 5 ;\ 15 ;\ 10 ;\ 25 ;\ 20 ;\ 35 ;\ 40 ;\ \ldots \] Upon inspection, the sequence alternates between two distinct patterns: 1. **Odd-Indexed Terms (Arithmetic Sequence):** - **Terms:** \( T_1 = 5 \), \( T_3 = 15 \), \( T_5 = 25 \), \( T_7 = 35 \), \(\ldots\) - **Pattern:** Each odd-indexed term increases by 10. - **General Formula:** \( T_n = 5n \) for odd \( n \). 2. **Even-Indexed Terms (Geometric Sequence):** - **Terms:** \( T_2 = 5 \), \( T_4 = 10 \), \( T_6 = 20 \), \( T_8 = 40 \), \(\ldots\) - **Pattern:** Each even-indexed term doubles the previous even term. - **General Formula:** \( T_n = 5 \times 2^{(n/2 - 1)} \) for even \( n \). ### 4.1 Write Down the Next 2 Terms Following the established patterns: - **Next Odd-Indexed Term (\( T_9 \)):** \[ T_9 = 5 \times 9 = 45 \] - **Next Even-Indexed Term (\( T_{10} \)):** \[ T_{10} = 5 \times 2^{(10/2 - 1)} = 5 \times 2^{4} = 5 \times 16 = 80 \] **Answer:** The next two terms are **45** and **80**.

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The observed sequence alternates between two patterns: the first subsequence (5, 15, 25, 35, ...) follows a linear increase of 10, while the second subsequence (5, 10, 20, 40, ...) follows a geometric progression with a common ratio of 2. Thus, the next two terms after 40 would be 45 (after 40, following the arithmetic pattern) and 80 (doubling 40, from the geometric pattern). For part 4.2, \( \mathrm{T}_{42} \) corresponds to the arithmetic pattern (the 42nd term in the sequence), while \( \mathrm{T}_{41} \) corresponds to the geometric segment. Since the patterns alternate, you can determine both values easily. The \( \mathrm{T}_{42} \) would thus be 45 and \( \mathrm{T}_{41} \) would be 80. Hence, \( \mathrm{T}_{42} - \mathrm{T}_{41} = 45 - 80 = -35 \). To find the sum of the first 50 terms in part 4.3, observe that the arithmetic sequence part has 25 terms (up to T50 are every second term), and forms an arithmetic series. The first term of this series is 5, and the last term at T50 (which is T25 of the arithmetic sequence) is 5 + (24 * 10) = 245. The formula for the sum of an arithmetic series \( S_n = \frac{n}{2} (a + l) \) where \( n = 25, a = 5, l = 245 \) gives us \( S_{25} = \frac{25}{2} (5 + 245) = 3125 \). The geometric part consists of the first 25 terms (5, 10, 20, …), which can be calculated using the geometric sum formula: \( S = a \frac{(r^n - 1)}{(r - 1)} \). To prove that all terms of the sequence are divisible by 5, we can express each term in the sequence as either \( 5n \) (for the arithmetic subsequence) or as \( 5 \times (2^m) \) (for the geometric subsequence). Since both representations show that every term is a multiple of 5, hence all terms in the sequence are divisible by 5.

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