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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.

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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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