ESTION 2 Consider the following quadratic sequence: \( 6 ; x ; 26 ; 45 ; y ; \ldots \) Determine the values of \( x \) and \( y \). 2 Given the following series: \( 220+213+206+\ldots-11 \) 2.2.1 Calculate the sum of the series. 2.2.2 Write the series in sigma-notation. 2.3 A batt is dropped from a beight of 15 m . It bounces back and loses \( 10 \% \) of its prev height on each bounce. Show that the total distance the ball will bounce cannot exi 290 m . 2.4 Given: \( 25\left(\frac{1-t}{3}\right)+5\left(\frac{1-1}{3}\right)^{2}+\left(\frac{1-1}{3}\right)^{3}+ \). \( \qquad \) 2.4.1 For which value(s) of \( l \) will the series converge? 2.4.2 If \( t=15 \), calculate the sum to infinity of the series if it exists. 2.5 The sum of the first \( n \) terms of a sequence is \( S_{\infty}=2^{x-1}+3 \). Deternine the \( 70^{\text {th }} \) term. Leave your answer in the form \( a . b^{\circ} \) where \( a, b \) and \( p \) at integers.

Let's break down the problem step by step. ### 1. Quadratic Sequence: \( 6 ; x ; 26 ; 45 ; y ; \ldots \) In a quadratic sequence, the second differences are constant. Let's denote the terms as follows: - \( a_1 = 6 \) - \( a_2 = x \) - \( a_3 = 26 \) - \( a_4 = 45 \) - \( a_5 = y \) First, we calculate the first differences: - \( d_1 = a_2 - a_1 = x - 6 \) - \( d_2 = a_3 - a_2 = 26 - x \) - \( d_3 = a_4 - a_3 = 45 - 26 = 19 \) - \( d_4 = a_5 - a_4 = y - 45 \) Now, we calculate the second differences: - \( d_2 - d_1 = (26 - x) - (x - 6) = 32 - 2x \) - \( d_3 - d_2 = 19 - (26 - x) = x - 7 \) - \( d_4 - d_3 = (y - 45) - 19 = y - 64 \) Setting the second differences equal to each other: 1. \( 32 - 2x = x - 7 \) 2. \( x - 7 = y - 64 \) Now, let's solve these equations. #### Solving the first equation: \[ 32 - 2x = x - 7 \] \[ 32 + 7 = 3x \] \[ 39 = 3x \implies x = 13 \] #### Solving the second equation: \[ 13 - 7 = y - 64 \] \[ 6 = y - 64 \implies y = 70 \] Thus, the values are: - \( x = 13 \) - \( y = 70 \) ### 2. Series: \( 220 + 213 + 206 + \ldots - 11 \) This is an arithmetic series where: - First term \( a = 220 \) - Common difference \( d = 213 - 220 = -7 \) #### 2.1 Calculate the sum of the series. To find the number of terms \( n \): \[ a_n = a + (n-1)d \] Setting \( a_n = -11 \): \[ -11 = 220 + (n-1)(-7) \] \[ -11 - 220 = -7(n-1) \] \[ -231 = -7(n-1) \implies n - 1 = \frac{231}{7} = 33 \implies n = 34 \] Now, we can calculate the sum \( S_n \): \[ S_n = \frac{n}{2} (a + a_n) = \frac{34}{2} (220 - 11) = 17 \times 209 = 3553 \] #### 2.2 Write the series in sigma-notation. The series can be expressed in sigma notation as: \[ \sum_{k=0}^{33} (220 - 7k) \] ### 2.3 Total Distance of the Bouncing Ball The ball is dropped from a height of \( 15 \) m and loses \( 10\% \) of its height on each bounce. The height after each bounce is: - First bounce: \( 15 \times 0.9 = 13.5 \) - Second bounce: \( 13.5 \times 0.9 = 12.15 \) - And so on... The total distance \( D \) can be calculated as: \[ D = 15 + 2 \sum_{n=1}^{\infty} 15 \times (0.9)^n \] The series \( \sum_{n=1}^{\infty} (0.9)^n \) is a geometric series with first term \( a = 0.9 \) and ratio \( r = 0.9 \): \[ \sum_{n=1}^{\infty} (0.9)^n = \frac{0.9}{1 - 0.9} = 9 \] Thus, \[ D = 15 + 2 \times 15 \times 9 = 15 + 270 = 285 \] Since \( 285 < 290 \), the total distance cannot exceed \( 290 \) m. ### 2.4 Series Convergence #### 2.4.1 For which value(s) of \( l \) will the series converge? The series converges if the common ratio \( r < 1 \). The common ratio is \( \frac{1 - l}{3} \): \[ \frac{1 - l}{3} < 1 \implies 1 - l < 3 \implies l > -2 \] #### 2.4.2 If \( t = 15 \), calculate the sum to infinity of the series if it exists. Substituting \( t = 15 \): \[ S = 25 \left(\frac{1 - 15}{3}\right) + 5 \left(\frac{1 - 1}{3}\right)^{2} + \left(\frac{1 - 1}{3}\right)^{3} + \ldots \] The first term dominates, and since the other terms vanish, we focus on: \[ S = 25 \left(\frac{-14}{3}\right) = -\frac{350}{3} \] ### 2.5 Determine the \( 70^{\text{th}} \) term Given \( S_{\infty} = 2^{x-1} + 3 \), the \( n^{\text{th}} \) term \( T_n \) can be expressed as: \[ T_n = S_n - S_{n-1} \] Assuming \( S_n = 2^{n-1} + 3 \): \[ S_{n-1} = 2^{n-2} + 3 \] Thus, \[ T_n = (2^{n-1} + 3) - (2^{n-2} + 3) = 2^{n-1} - 2^{n-2} = 2^{n-2} \] For \( n = 70 \): \[ T_{70} = 2^{70-2} = 2^{68} \] The answer in the form \( a \cdot b^{p} \) is: \[ 1 \cdot 2^{68} \] ### Summary of Answers: - \( x = 13 \), \( y = 70 \