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: $$ 25\left(\frac{1-t}{3} ight)+5\left(\frac{1-1}{3} ight)^{2}+\left(\frac{1-1}{3} ight)^{3}+ $$. $$ \qquad $$ 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 $$ 70^{ ext {th }} $$ term. Leave your answer in the form $$ a . b^{\circ} $$ where $$ a, b $$ and $$ p $$ at integers.

Question

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: $$ 25\left(\frac{1-t}{3}ight)+5\left(\frac{1-1}{3}ight)^{2}+\left(\frac{1-1}{3}ight)^{3}+ $$. $$ \qquad $$ 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 $$ 70^{	ext {th }} $$ term. Leave your answer in the form $$ a . b^{\circ} $$ where $$ a, b $$ and $$ p $$ at integers.

UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### 1. Quadratic Sequence: $$ 6 ; x ; 26 ; 45 ; y ; \ldots $$

A quadratic sequence has a constant second difference. 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:
- $$ a_2 - a_1 = x - 6 $$
- $$ a_3 - a_2 = 26 - x $$
- $$ a_4 - a_3 = 45 - 26 = 19 $$
- $$ a_5 - a_4 = y - 45 $$

Now, the first differences are:
- $$ d_1 = x - 6 $$
- $$ d_2 = 26 - x $$
- $$ d_3 = 19 $$
- $$ d_4 = y - 45 $$

Next, we calculate the second differences:
- $$ d_2 - d_1 = (26 - x) - (x - 6) = 32 - 2x $$
- $$ d_3 - d_2 = 19 - (26 - x) = x - 7 $$
- $$ d_4 - d_3 = (y - 45) - 19 = y - 64 $$

Since the second differences must be constant, we set them equal:
1. $$ 32 - 2x = x - 7 $$
2. $$ x - 7 = y - 64 $$

Let's solve these equations step by step.

#### 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. Given the 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 $$ n $$, we use the formula for the $$ n $$-th term of an arithmetic sequence:
$$
a_n = a + (n-1)d
$$
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 $$ S_n $$ of the first $$ n $$ terms:
$$
S_n = \frac{n}{2} (a + a_n) = \frac{34}{2} (220 - 11) = 17 \times 209 = 3553
$$

#### 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 A ball dropped from a height of 15 m.

The ball loses $$ 10\% $$ of its height on each bounce, meaning it retains $$ 90\% $$ of its height. The heights of the bounces form a geometric series:
- First height $$ h_1 = 15 $$
- Common ratio $$ r = 0.9 $$

The total distance $$ D $$ is given by:
$$
D = h_1 + 2(h_2 + h_3 + h_4 + \ldots)
$$
Where $$ h_2 = 15 \times 0.9 $$, $$ h_3 = 15 \times 0.9^2 $$, etc.

The sum of the heights of the bounces is:
$$
h_2 + h_3 + h_4 + \ldots = 15 \times 0.9 \left( \frac{1}{1 - 0.9} \right) = 15 \times 0.9 \times 10 = 135
$$

Thus, the total distance is:
$$
D = 15 + 2 \times 135 = 15 + 270 = 285 \text{ m}
$$

Since $$ 285 < 290 $$, the total distance cannot exceed $$ 290 $$ m.

### 2.4 Given series: $$ 25\left(\frac{1-t}{3}\right)+5\left(\frac{1-1}{3}\right)^{2}+\left(\frac{1-1}{3}\right)^{3}+ $$

#### 2.4.1 For which value(s) of $$ l $$ will the series converge?

This series appears to be a geometric series. The series converges if the common ratio $$ r < 1 $$. Here, we need to analyze the term $$ \frac{1 - t}{3} $$. For convergence:
$$
\left| \frac{1 - t}{3} \right| < 1 \implies -1 < \frac{1 - t}{3} < 1
$$
This leads to:
$$
-3 < 1 - t < 3 \implies -2 < -t < 4 \implies -4 < t < 2
$$

#### 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.

### 2.5 The sum of the first $$ n $$ terms of a sequence is $$ S_{\infty} = 2^{x-1} + 3 $$.

To find the $$ 70^{\text{th}} $$ term, we need to express the $$ n $$-th term $$ T_n $$:
$$
T_n = S_n - S_{n-1}
$$
Given $$ S_n = 2^{n-1} + 3 $$:
$$
S_{n-1} = 2^{(n-1)-1} + 3 = 2^{n-2} + 3
$$
Thus,
\[
T_n = (2^{n-1} + 3) - (2^{n-2} + 3)
The 70th term is $$ 2^{69} $$.

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