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10 terms is 455 . Calculate the first three terms. Determine three consecutive terms of an arithmetic sequence. if the middle term is \( r \), the sum of the three terms is 27 and the product is 648 .

Ask by Riley Vega. in South Africa
Jan 23,2025

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Answer

Cannot determine the first three terms uniquely with the given information. For the second problem, the three consecutive terms are 6, 9, and 12.

Solution

Let's tackle your questions step-by-step. It appears there are two separate problems: 1. **Calculating the First Three Terms Given the Sum of 10 Terms** 2. **Determining Three Consecutive Terms of an Arithmetic Sequence with Specific Conditions** --- ## 1. Calculating the First Three Terms Given the Sum of 10 Terms **Problem Statement:** The sum of the first 10 terms of an arithmetic sequence is 455. Calculate the first three terms. **Solution:** To determine the first three terms of an arithmetic sequence where the sum of the first 10 terms (S₁₀) is 455, we'll use the formula for the sum of an arithmetic sequence: \[ S_n = \frac{n}{2} [2a + (n - 1)d] \] Where: - \( S_n \) = Sum of the first \( n \) terms - \( a \) = First term - \( d \) = Common difference - \( n \) = Number of terms Given: \[ S_{10} = 455 \] \[ n = 10 \] Plugging the values into the formula: \[ 455 = \frac{10}{2} [2a + 9d] \] \[ 455 = 5(2a + 9d) \] \[ 2a + 9d = 91 \quad \text{(Equation 1)} \] **Challenge:** With only the sum of 10 terms provided, there are infinitely many possible solutions for \( a \) and \( d \) because we have one equation with two unknowns. **Conclusion:** Without additional information (such as another term in the sequence or the common difference), we cannot uniquely determine the first three terms of the sequence. More data is needed to find specific values for \( a \) and \( d \). --- ## 2. Determining Three Consecutive Terms of an Arithmetic Sequence **Problem Statement:** Determine three consecutive terms of an arithmetic sequence. If the middle term is \( r \), the sum of the three terms is 27, and the product is 648. **Solution:** Let’s denote the three consecutive terms of the arithmetic sequence as: \[ (r - d), \quad r, \quad (r + d) \] Where: - \( r \) = Middle term - \( d \) = Common difference **Given:** 1. **Sum of the three terms is 27:** \[ (r - d) + r + (r + d) = 27 \] \[ 3r = 27 \] \[ r = 9 \] 2. **Product of the three terms is 648:** \[ (r - d) \times r \times (r + d) = 648 \] \[ (9 - d) \times 9 \times (9 + d) = 648 \] Simplify: \[ 9 \times (81 - d^2) = 648 \] \[ 729 - 9d^2 = 648 \] \[ 9d^2 = 729 - 648 \] \[ 9d^2 = 81 \] \[ d^2 = 9 \] \[ d = \pm 3 \] **Possible Common Differences:** - \( d = 3 \) - \( d = -3 \) **Calculating the Terms:** 1. **If \( d = 3 \):** \[ \text{First term: } 9 - 3 = 6 \] \[ \text{Second term: } 9 \] \[ \text{Third term: } 9 + 3 = 12 \] \[ \text{Sequence: } 6, \; 9, \; 12 \] 2. **If \( d = -3 \):** \[ \text{First term: } 9 - (-3) = 12 \] \[ \text{Second term: } 9 \] \[ \text{Third term: } 9 + (-3) = 6 \] \[ \text{Sequence: } 12, \; 9, \; 6 \] **Conclusion:** Both sequences \( 6, \; 9, \; 12 \) and \( 12, \; 9, \; 6 \) satisfy the given conditions. Typically, we present the sequence in increasing order, so: \[ \boxed{6,\ 9,\ 12} \] --- **Summary:** 1. **First Problem:** Cannot determine the first three terms uniquely with the given information. 2. **Second Problem:** The three consecutive terms are **6, 9, and 12**. If you have more information for the first problem or need further assistance, feel free to ask!

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The Deep Dive

To find the first three terms of an arithmetic sequence where the sum of 10 terms is 455, we can start by determining the average of those terms. The average would be \( \frac{455}{10} = 45.5 \). Each term can then be expressed as \( 45.5 - 4d \), \( 45.5 \), and \( 45.5 + 4d \) for \( d \) being the common difference. By solving, we’ll conclude that the first three terms can vary depending on the value of \( d \), but will revolve around 45.5. For the subsequent problem, if the middle term is \( r \) in the three-term sequence, we set the terms as \( r-1d \), \( r \), and \( r+d \). We know that the sum is given by \( 3r = 27 \) or \( r = 9 \). The product can be expressed as \( (r - d)r(r + d) = 648 \). Plugging \( r = 9 \) into this gives us \( (9 - d)9(9 + d) = 648 \), allowing us to solve for \( d \). This yields interesting integer solutions for \( d \), leading to three specific consecutive terms!

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