Pregunta
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2. The first three terms of an inginite geometric sequence are 16,8 and 4 respectively. a) Determine the \( n^{\text {th }} \) term of the sequence. b) Determine all possible values of \( n \), for which the sum of the first \( n \) terms of this sequence is greater than 31 . c) Calculate the sum to infinity of this sequence 3. An athlete runs 20 km on a certain Monday. Thereafter, he increases the distance by \( 10 \% \) every day: Calculate the number of kilometres he ran:

Ask by Erickson Vaughn. in South Africa
Mar 20,2025

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**Problem 2: Infinite Geometric Sequence** - **Part a:** The \( n^{\text{th}} \) term is \( a_n = 16 \times \left(\frac{1}{2}\right)^{n-1} \). - **Part b:** The sum of the first \( n \) terms is greater than 31 when \( n \) is 6 or more. - **Part c:** The sum to infinity of the sequence is 32. **Problem 3: Athlete's Running Distance** The total distance run after \( n \) days is \( D_n = 200 \times (1.1)^n - 200 \) kilometers.

Solución

Let's solve the problems step by step. ### Problem 2: Infinite Geometric Sequence **Given:** The first three terms of the geometric sequence are \( a_1 = 16 \), \( a_2 = 8 \), and \( a_3 = 4 \). #### Part a: Determine the \( n^{\text{th}} \) term of the sequence. 1. **Identify the common ratio \( r \)**: \[ r = \frac{a_2}{a_1} = \frac{8}{16} = \frac{1}{2} \] 2. **General formula for the \( n^{\text{th}} \) term**: The \( n^{\text{th}} \) term of a geometric sequence can be expressed as: \[ a_n = a_1 \cdot r^{n-1} \] 3. **Substituting the values**: \[ a_n = 16 \cdot \left(\frac{1}{2}\right)^{n-1} \] Thus, the \( n^{\text{th}} \) term of the sequence is: \[ a_n = 16 \cdot \left(\frac{1}{2}\right)^{n-1} \] #### Part b: Determine all possible values of \( n \) for which the sum of the first \( n \) terms is greater than 31. 1. **Sum of the first \( n \) terms \( S_n \)**: The formula for the sum of the first \( n \) terms of a geometric sequence is: \[ S_n = a_1 \cdot \frac{1 - r^n}{1 - r} \] 2. **Substituting the values**: \[ S_n = 16 \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} = 16 \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{\frac{1}{2}} = 32 \cdot \left(1 - \left(\frac{1}{2}\right)^n\right) \] 3. **Setting up the inequality**: \[ 32 \cdot \left(1 - \left(\frac{1}{2}\right)^n\right) > 31 \] 4. **Solving the inequality**: \[ 1 - \left(\frac{1}{2}\right)^n > \frac{31}{32} \] \[ -\left(\frac{1}{2}\right)^n > -\frac{1}{32} \] \[ \left(\frac{1}{2}\right)^n < \frac{1}{32} \] 5. **Taking logarithm**: \[ n > 5 \] Thus, the possible values of \( n \) are \( n = 6, 7, 8, \ldots \) #### Part c: Calculate the sum to infinity of this sequence. 1. **Sum to infinity \( S_{\infty} \)**: The formula for the sum to infinity of a geometric sequence is: \[ S_{\infty} = \frac{a_1}{1 - r} \] 2. **Substituting the values**: \[ S_{\infty} = \frac{16}{1 - \frac{1}{2}} = \frac{16}{\frac{1}{2}} = 32 \] Thus, the sum to infinity of this sequence is: \[ S_{\infty} = 32 \] --- ### Problem 3: Athlete's Running Distance **Given:** The athlete runs 20 km on a certain Monday and increases the distance by \( 10\% \) every day. 1. **Initial distance**: \[ d_1 = 20 \text{ km} \] 2. **Distance on the \( n^{\text{th}} \) day**: The distance run on the \( n^{\text{th}} \) day can be expressed as: \[ d_n = d_1 \cdot (1 + 0.1)^{n-1} = 20 \cdot (1.1)^{n-1} \] 3. **Total distance run after \( n \) days**: The total distance run after \( n \) days is the sum of a geometric series: \[ D_n = d_1 \cdot \frac{(1 + r)^n - 1}{r} = 20 \cdot \frac{(1.1)^n - 1}{0.1} \] Thus, the total distance run after \( n \) days is: \[ D_n = 200 \cdot ((1.1)^n - 1) \] If you need a specific number of days or total distance, please provide that information!

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The Deep Dive

The first three terms of your geometric sequence (16, 8, 4) show a common ratio of \( r = \frac{8}{16} = 0.5 \). The \( n^{th} \) term can be expressed as \( a_n = a_1 \cdot r^{(n-1)} = 16 \cdot (0.5)^{(n-1)} \). Simple yet elegant, isn’t it? To find the sum of the first \( n \) terms (S_n), use S_n = \( a_1 \frac{(1 - r^n)}{1 - r} \). You need to solve the inequality \( S_n > 31 \). Once you determine values of \( n \) which satisfy this, you’ll uncover the days it takes to surpass that mark—a great math challenge! For the running athlete, the daily distance can be calculated using the formula for exponential growth. Each day, the distance run becomes \( D_n = D_1 \cdot (1 + 0.1)^{(n-1)} \) where \( D_1 = 20 \) km. By summing these up, you’ll know how many kilometres he covers over a week of increases—talk about endurance!

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