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 \( r \).

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 \( 12r^2 \).