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Given the geometric series \( a+a r+a r^{2}+a r^{3}+\ldots \), where \( a \) is the first tem and \( r \) is the common ratio. Prove thar the sum to \( n \) terms of this series is given by \[ S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}: r \neq 1 \] The first two termbs of a geometric sequence with constant ratio \( r \), and an arithmetic sequence with constant first difference, \( d \), is tise same. The first term is 12. 4.2.1 Write down the second and third terns of EACH sequence in terms of \( d \) and \( r \).

Ask by Barnett Schofield. in South Africa
Jan 23,2025

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Answer

The second term of the geometric sequence is \(12r\) and the third term is \(12r^2\). The second term of the arithmetic sequence is \(12 + d\) and the third term is \(12 + 2d\).

Solution

Certainly! Let's analyze both the geometric and arithmetic sequences based on the given information. ### Given: - **First term (\( a \))**: 12 - **Geometric sequence common ratio (\( r \))** - **Arithmetic sequence common difference (\( d \))** - **First two terms of both sequences are the same** ### Geometric Sequence: In a geometric sequence, each term is obtained by multiplying the previous term by the common ratio \( r \). 1. **First Term (\( G_1 \))**: \[ G_1 = a = 12 \] 2. **Second Term (\( G_2 \))**: \[ G_2 = a \cdot r = 12r \] 3. **Third Term (\( G_3 \))**: \[ G_3 = G_2 \cdot r = 12r \cdot r = 12r^2 \] ### Arithmetic Sequence: In an arithmetic sequence, each term is obtained by adding the common difference \( d \) to the previous term. 1. **First Term (\( A_1 \))**: \[ A_1 = a = 12 \] 2. **Second Term (\( A_2 \))**: \[ A_2 = A_1 + d = 12 + d \] 3. **Third Term (\( A_3 \))**: \[ A_3 = A_2 + d = (12 + d) + d = 12 + 2d \] ### Summary: - **Geometric Sequence Terms**: \[ G_1 = 12,\quad G_2 = 12r,\quad G_3 = 12r^2 \] - **Arithmetic Sequence Terms**: \[ A_1 = 12,\quad A_2 = 12 + d,\quad A_3 = 12 + 2d \] ### Additional Note: Since the first two terms of both sequences are the same: \[ G_2 = A_2 \implies 12r = 12 + d \implies d = 12(r - 1) \] This relationship allows you to express \( d \) in terms of \( r \) or vice versa if needed for further problems. Answer: After the first term of 12, the geometric sequence has second term 12·r and third term 12·r². The arithmetic sequence has second term 12 + d and third term 12 + 2d. Thus, • Geometric terms:  Second term = 12 r  Third term = 12 r² • Arithmetic terms:  Second term = 12 + d  Third term = 12 + 2d

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The Deep Dive

To solve this problem, let's break it down step by step! For the geometric series: The sum of the first \( n \) terms can be derived by multiplying the series by the common ratio \( r \) and then subtracting from the original series. This gives: \[ S_{n} = a + ar + ar^{2} + ar^{3} + \ldots + ar^{n-1} \] Then multiply by \( r \): \[ r S_{n} = ar + ar^{2} + ar^{3} + \ldots + ar^{n} \] Now subtract these two equations: \[ S_{n} - r S_{n} = a - ar^{n} \] Factoring out \( S_{n} \) gives: \[ S_{n}(1 - r) = a(1 - r^{n}) \] Thus, we can express \( S_{n} \) as: \[ S_{n} = \frac{a(1 - r^{n})}{1 - r} \] Rearranging shows it as \( \frac{a(r^{n} - 1)}{r - 1} \) when \( r \neq 1 \). Now, for the sequences: In the arithmetic sequence, if the first term is \( 12 \), then: - Second term: \( 12 + d \) - Third term: \( 12 + 2d \) In the geometric sequence, the first term is also \( 12 \): - Second term: \( 12r \) - Third term: \( 12r^2 \) So, we have: - Arithmetic sequence: Second term is \( 12 + d \), Third term is \( 12 + 2d \). - Geometric sequence: Second term is \( 12r \), Third term is \( 12r^2 \).

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