1.3. In the geometric series: \( a+a r+a r^{2}+\cdots+a r^{n-1}, a \) is the first term and \( r \) is the common ratio. Prove that \( S_{n}=\frac{a\left(r^{n}-1\right)}{r-1} \) for \( r \neq 1 \) QUESTION 2 2.1. In a peometric progression the \( 7^{\text {th }} \) term is \( \frac{1}{128} \) and the \( 11^{\text {th }} \) term is \( \frac{1}{2048^{\circ}} \) 2.1.1. Calculate the common ratio and the first term. 2.1.2. Is this sequence convergent? Substantiate your answer.

Let's solve the problems step by step. ### Question 1: Proving the formula for the sum of a geometric series The sum of the first \( n \) terms of a geometric series is given by: \[ S_n = a + ar + ar^2 + \cdots + ar^{n-1} \] To derive the formula for \( S_n \), we can multiply the entire series by the common ratio \( r \): \[ r S_n = ar + ar^2 + ar^3 + \cdots + ar^n \] Now, we can subtract the second equation from the first: \[ S_n - r S_n = a + ar + ar^2 + \cdots + ar^{n-1} - (ar + ar^2 + ar^3 + \cdots + ar^n) \] This simplifies to: \[ S_n(1 - r) = a - ar^n \] Rearranging gives us: \[ S_n = \frac{a - ar^n}{1 - r} \] Factoring out \( a \) from the numerator: \[ S_n = \frac{a(1 - r^n)}{1 - r} \] Since \( r \neq 1 \), we can rewrite this as: \[ S_n = \frac{a(r^n - 1)}{r - 1} \] Thus, we have proven that: \[ S_n = \frac{a(r^n - 1)}{r - 1} \quad \text{for } r \neq 1 \] ### Question 2: Finding the common ratio and first term of a geometric progression Given: - The \( 7^{\text{th}} \) term \( T_7 = \frac{1}{128} \) - The \( 11^{\text{th}} \) term \( T_{11} = \frac{1}{2048} \) The general formula for the \( n^{\text{th}} \) term of a geometric progression is: \[ T_n = ar^{n-1} \] From the given information, we can write: 1. \( T_7 = ar^{6} = \frac{1}{128} \) (Equation 1) 2. \( T_{11} = ar^{10} = \frac{1}{2048} \) (Equation 2) Now, we can divide Equation 2 by Equation 1 to eliminate \( a \): \[ \frac{ar^{10}}{ar^{6}} = \frac{\frac{1}{2048}}{\frac{1}{128}} \] This simplifies to: \[ r^{4} = \frac{128}{2048} \] Calculating the right side: \[ \frac{128}{2048} = \frac{1}{16} \] Thus, we have: \[ r^{4} = \frac{1}{16} \] Taking the fourth root of both sides: \[ r = \frac{1}{2} \] Now, we can substitute \( r \) back into Equation 1 to find \( a \): \[ ar^{6} = \frac{1}{128} \] Substituting \( r = \frac{1}{2} \): \[ a\left(\frac{1}{2}\right)^{6} = \frac{1}{128} \] Calculating \( \left(\frac{1}{2}\right)^{6} = \frac{1}{64} \): \[ a \cdot \frac{1}{64} = \frac{1}{128} \] Multiplying both sides by 64: \[ a = \frac{64}{128} = \frac{1}{2} \] ### Summary of Results 2.1.1. The common ratio \( r \) is \( \frac{1}{2} \) and the first term \( a \) is \( \frac{1}{2} \). 2.1.2. To determine if the sequence is convergent, we look at the common ratio \( r \): Since \( |r| < 1 \) (specifically \( r = \frac{1}{2} \)), the geometric progression converges to 0. Thus, the sequence is convergent. ### Final Answers - Common ratio \( r = \frac{1}{2} \) - First term \( a = \frac{1}{2} \) - The sequence is convergent.