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. ### Part 1: Quadratic Sequence The sequence given is \( 6, x, 26, 45, y, \ldots \). 1. **Identify the differences:** - First difference: \( x - 6 \) - Second difference: \( 26 - x \) - Third difference: \( 45 - 26 = 19 \) - Fourth difference: \( y - 45 \) 2. **Calculate the first differences:** - First difference: \( d_1 = x - 6 \) - Second difference: \( d_2 = 26 - x \) - Third difference: \( d_3 = 19 \) - Fourth difference: \( d_4 = y - 45 \) 3. **Set up the equations based on the differences:** - The second differences should be constant for a quadratic sequence. - Thus, \( d_2 - d_1 = d_3 \) and \( d_3 - d_2 = d_4 \). 4. **Solve for \( x \) and \( y \):** - From \( d_2 - d_1 = 19 \): \[ (26 - x) - (x - 6) = 19 \] \[ 32 - 2x = 19 \implies 2x = 13 \implies x = 6.5 \] - From \( d_3 - d_2 = d_4 \): \[ 19 - (26 - x) = y - 45 \] Substituting \( x = 6.5 \): \[ 19 - (26 - 6.5) = y - 45 \] \[ 19 - 19.5 = y - 45 \implies -0.5 = y - 45 \implies y = 44.5 \] Thus, the values are \( x = 6.5 \) and \( y = 44.5 \). ### Part 2: Series Calculation The series is \( 220 + 213 + 206 + \ldots - 11 \). #### 2.2.1 Calculate the sum of the series. 1. **Identify the first term and common difference:** - First term \( a = 220 \) - Common difference \( d = 213 - 220 = -7 \) 2. **Find the number of terms:** - The last term is \( -11 \). - Use the formula for the \( n \)-th term of an arithmetic series: \[ a_n = a + (n-1)d \] Setting \( a_n = -11 \): \[ -11 = 220 + (n-1)(-7) \] \[ -11 - 220 = -7(n-1) \implies -231 = -7(n-1) \implies n-1 = 33 \implies n = 34 \] 3. **Calculate the sum:** - Use the sum formula for an arithmetic series: \[ S_n = \frac{n}{2} (a + a_n) \] \[ S_{34} = \frac{34}{2} (220 - 11) = 17 \times 209 = 3553 \] #### 2.2.2 Write the series in sigma-notation. The series can be expressed in sigma notation as: \[ \sum_{k=0}^{33} (220 - 7k) \] ### Part 3: Bouncing Ball A ball is dropped from a height of \( 15 \) m and loses \( 10\% \) of its height on each bounce. 1. **Calculate the total distance:** - The height after each bounce is \( 90\% \) of the previous height. - The heights form a geometric series: - First drop: \( 15 \) - First bounce: \( 15 \times 0.9 = 13.5 \) - Second drop: \( 13.5 \) - Second bounce: \( 13.5 \times 0.9 = 12.15 \) - And so on... 2. **Total distance:** - The total distance \( D \) is: \[ D = 15 + 2(13.5 + 12.15 + \ldots) \] - The series \( 13.5 + 12.15 + \ldots \) is a geometric series with first term \( a = 13.5 \) and common ratio \( r = 0.9 \). - The sum of an infinite geometric series is: \[ S = \frac{a}{1 - r} = \frac{13.5}{1 - 0.9} = \frac{13.5}{0.1} = 135 \] - Thus, the total distance: \[ D = 15 + 2 \times 135 = 15 + 270 = 285 \] The total distance the ball will bounce cannot exceed \( 290 \) m. ### Part 4: Series Convergence Given: \[ 25\left(\frac{1-t}{3}\right) + 5\left(\frac{1-1}{3}\right)^{2} + \left(\frac{1-1}{3}\right)^{3} + \ldots \] #### 2.4.1 For which value(s) of \( t \) will the series converge? The series converges if the common ratio \( r < 1 \): \[ \left|\frac{1-t}{3}\right| < 1 \] This leads to: \[ -1 < \frac{1-t}{3} < 1 \] Multiplying by \( 3 \): \[ -3 < 1 - t < 3 \] This gives: \[ -2 < -t < 4 \implies -4 < t < 2 \] #### 2.4.2 If \( t = 15 \), calculate the sum to infinity of the series if it exists. Since \( t = 15 \) is outside the convergence range, the series does not converge. ### Part 5: Sum of the First \( n \) Terms Given: \[ S_{\infty} = 2^{x-1} + 3 \] To find the \( 70^{th} \) term, we need to express the \( n \)-th term \( T_n \): \[ T_n = S_n - S_{n-1} \] Assuming \( S_n = 2^{n-1}