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. (2) 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:

Ask by Kelly Carroll. in South Africa
Mar 20,2025

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**Problem 2: Geometric Sequence** a) The \( n^{\text{th}} \) term is \( a_n = 16 \cdot \left(\frac{1}{2}\right)^{n-1} \). b) The sum of the first \( n \) terms is greater than 31 when \( n > 5 \). c) The sum to infinity is 32. **Problem 3: Athlete's Running Distance** The distance on the \( n^{\text{th}} \) day is \( d_n = 20 \cdot (1.1)^{n-1} \).

Solución

Let's solve the problems step by step. ### Problem 2: Geometric Sequence **Given:** The first three terms of the geometric sequence are \( a_1 = 16 \), \( a_2 = 8 \), and \( a_3 = 4 \). #### a) Determine the \( n^{\text{th}} \) term of the sequence. In a geometric sequence, the \( n^{\text{th}} \) term can be expressed as: \[ a_n = a_1 \cdot r^{n-1} \] where \( r \) is the common ratio. To find \( r \), we can use the first two terms: \[ r = \frac{a_2}{a_1} = \frac{8}{16} = \frac{1}{2} \] Now, we can express the \( n^{\text{th}} \) term: \[ a_n = 16 \cdot \left(\frac{1}{2}\right)^{n-1} \] #### b) Determine all possible values of \( n \) for which the sum of the first \( n \) terms of this sequence is greater than 31. The sum of the first \( n \) terms of a geometric sequence is given by: \[ S_n = a_1 \cdot \frac{1 - r^n}{1 - r} \] Substituting the known values: \[ S_n = 16 \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} = 16 \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{\frac{1}{2}} = 32 \cdot \left(1 - \left(\frac{1}{2}\right)^n\right) \] We need to find \( n \) such that: \[ 32 \cdot \left(1 - \left(\frac{1}{2}\right)^n\right) > 31 \] Dividing both sides by 32: \[ 1 - \left(\frac{1}{2}\right)^n > \frac{31}{32} \] Rearranging gives: \[ -\left(\frac{1}{2}\right)^n > -\frac{1}{32} \] \[ \left(\frac{1}{2}\right)^n < \frac{1}{32} \] Since \( \frac{1}{32} = \left(\frac{1}{2}\right)^5 \), we have: \[ n > 5 \] Thus, the possible values of \( n \) are \( n = 6, 7, 8, \ldots \) #### c) Calculate the sum to infinity of this sequence. The sum to infinity \( S \) of a geometric series is given by: \[ S = \frac{a_1}{1 - r} \] Substituting the known values: \[ S = \frac{16}{1 - \frac{1}{2}} = \frac{16}{\frac{1}{2}} = 32 \] ### Problem 3: Athlete's Running Distance **Given:** The athlete runs 20 km on a certain Monday and increases the distance by \( 10\% \) every day. The distance run on the \( n^{\text{th}} \) day can be expressed as: \[ d_n = 20 \cdot (1.1)^{n-1} \] If you have any specific questions or further calculations regarding this problem, please let me know!

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The Deep Dive

In an infinite geometric sequence, the ratio between successive terms is constant. From your sequence, the ratio is obtained by dividing the second term by the first (i.e., \( 8/16 = 0.5 \)). Therefore, the \( n^{th} \) term can be expressed as \( a_n = 16 \cdot (0.5)^{n-1} \). This means you have a clear formula to plug in any \( n \) to find the term you need! For sum calculations, when it comes to finding the sum of the first \( n \) terms of a geometric series, the formula is \( S_n = a \frac{1 - r^n}{1 - r} \). You'll need to solve \( S_n > 31 \) using the known values of the first term (16) and common ratio (0.5). Just keep in mind the exponential nature of the growth, and before you know it, you'll uncover all possible values for \( n \)!

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