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:
- $$ d_1 = a_2 - a_1 = x - 6 $$
- $$ d_2 = a_3 - a_2 = 26 - x $$
- $$ d_3 = a_4 - a_3 = 45 - 26 = 19 $$
- $$ d_4 = a_5 - a_4 = y - 45 $$

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

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

Let's solve these equations to find $$ x $$ and $$ y $$.

### 2. Series: $$ 220 + 213 + 206 + \ldots - 11 $$

This is an arithmetic series where:
- First term $$ a = 220 $$
- Common difference $$ d = 213 - 220 = -7 $$

To find the number of terms $$ n $$ until we reach -11, we use the formula for the $$ n $$-th term of an arithmetic sequence:
$$
a_n = a + (n-1)d
$$
Setting $$ a_n = -11 $$:
$$
-11 = 220 + (n-1)(-7)
$$
Now, let's solve for $$ n $$.

### 3. Ball Bouncing Problem

The ball is dropped from a height of 15 m and loses 10% of its height on each bounce. The height after each bounce forms a geometric series:
- First height $$ h_1 = 15 $$
- Each subsequent height $$ h_n = h_1 \cdot (0.9)^{n-1} $$

The total distance traveled by the ball is the sum of the distances falling and bouncing back:
$$
D = h_1 + 2(h_1 \cdot 0.9 + h_1 \cdot 0.9^2 + h_1 \cdot 0.9^3 + \ldots)
$$
This is a geometric series. Let's calculate the total distance.

### 4. Series Convergence

The series given is:
$$
25\left(\frac{1-t}{3}\right) + 5\left(\frac{1-1}{3}\right)^{2} + \left(\frac{1-1}{3}\right)^{3} + \ldots
$$
We need to determine for which values of $$ l $$ the series converges.

### 5. Sum of First $$ n $$ Terms

Given:
$$
S_{\infty} = 2^{x-1} + 3
$$
We need to find the $$ 70^{th} $$ term. The $$ n $$-th term can be expressed as:
$$
T_n = S_n - S_{n-1}
$$
Let's find $$ T_{70} $$.

Now, I will proceed to solve the equations and calculations step by step.

### Step 1: Solve for $$ x $$ and $$ y $$

Let's solve the equations:
1. $$ 32 - 2x = x - 7 $$
2. $$ x - 7 = y - 64 $$

Starting with the first equation:
$$
32 - 2x = x - 7 \implies 32 + 7 = 3x \implies 39 = 3x \implies x = 13
$$

Now substituting $$ x $$ into the second equation:
$$
13 - 7 = y - 64 \implies 6 = y - 64 \implies y = 70
$$

Thus, $$ x = 13 $$ and $$ y = 70 $$.

### Step 2: Calculate the sum of the series

Now, let's find $$ n $$ for the series $$ 220 + 213 + 206 + \ldots - 11 $$:
$$
-11 = 220 + (n-1)(-7) \implies -11 - 220 = -7(n-1) \implies -231 = -7(n-1) \implies n-1 = 33 \implies n = 34
$$

The number of terms is $$ n = 34 $$.

Now, the sum $$ S_n $$ of an arithmetic series is given by:
$$
S_n = \frac{n}{2} (a + l)
$$
where $$ l $$ is the last term. Here, $$ l = -11 $$:
$$
S_{34} = \frac{34}{2} (220 - 11) = 17 \cdot 209 = 3553
$$

### Step 3: Total distance the ball will bounce

The total distance $$ D $$ is:
$$
D = 15 + 2 \left( 15 \cdot \frac{0.9}{1 - 0.9} \right) = 15 + 2 \left( 15 \cdot 9 \right) = 15 + 270 = 285
$$
Thus, the total distance is less than 290 m.

### Step 4: Series convergence

The series converges if the common ratio $$ r < 1 $$. Here, $$ r = \frac{1-1}{3} = 0 $$, which converges for all $$ l $$.

### Step 5: Find $$ T_{70} $$

To find $$ T_{70} $$:
$$
S_{70} = 2^{70-1} + 3 = 2^{69} + 3
$$
$$
S_{69} = 2^{69-1} + 3 = 2^{68} + 3
$$
Thus,
$$
T_{70} = S_{70} - S_{69} = (2^{69} + 3) - (2^{68} + 3) = 2^{69} - 2^{68} = 2^{68}
$$

### Final Answers:
- $$ x = 13 $$
-
$$ x = 13 $$ and $$ y = 70 $$.

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