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.

### 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:**
   - Since the second differences are constant in a quadratic sequence, we have:
     $$
     d_2 - d_1 = d_3
     $$
     $$
     (26 - x) - (x - 6) = 19
     $$
     Simplifying gives:
     $$
     32 - 2x = 19 \implies 2x = 13 \implies x = 6.5
     $$

4. **Now find $$ y $$:**
   - Using the same logic:
     $$
     d_4 - d_3 = d_3
     $$
     $$
     (y - 45) - 19 = 19
     $$
     Simplifying gives:
     $$
     y - 64 = 19 \implies y = 83
     $$

Thus, $$ x = 6.5 $$ and $$ y = 83 $$.

### 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 $$ a $$ and the common difference $$ d $$:**
   - First term $$ a = 220 $$
   - Common difference $$ d = 213 - 220 = -7 $$

2. **Find the number of terms $$ n $$:**
   - The last term is $$ -11 $$.
   - Using 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 $$ S_n $$:**
   $$
   S_n = \frac{n}{2} (a + a_n) = \frac{34}{2} (220 - 11) = 17 \times 209 = 3553
   $$

#### 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 $$ 15 \times (0.9)^n $$.
   - The total distance $$ D $$ is:
     $$
     D = 15 + 2 \sum_{n=1}^{\infty} 15 \times (0.9)^n
     $$
   - The series is a geometric series with first term $$ 15 \times 0.9 $$ and ratio $$ 0.9 $$:
     $$
     \sum_{n=1}^{\infty} 15 \times (0.9)^n = \frac{15 \times 0.9}{1 - 0.9} = 15 \times 9 = 135
     $$
   - 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 $$ r < 1 $$:
$$
\left|\frac{1-t}{3}\right| < 1 \implies -1 < \frac{1-t}{3} < 1
$$
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.

### Part 5: Sum of the First $$ n $$ Terms

Given:
$$
S_{\infty} = 2^{x-1} + 3
$$

To find the $$ 70^{th} $$ term $$ T_n $$:
1. The sum of the first $$ n $$ terms is given by:
   $$
   S_n = T_1 + T_2 + \ldots + T_n
   $$
2. The $$ n $$-th term can be expressed as:
   $$
   T_n = S_n - S_{n-1}
   $$
3. We need to find $$ S_{70} $$ and $$ S_{69} $$:
   $$
   S_{70} = 2^{70-1} + 3 = 2^{69} + 3
   $$
   $$
   S_{69} = 2^{69-1} + 3 = 2^{68} + 3
   $$
4. Thus:
   $$
   T_{70} = S_{70} - S_{69} = (2^{69} + 3) - (2^{68} + 3) = 2^{69} - 2^{68} = 2^{68}
   $$

The $$ 70^{th} $$ term is $$ 
$$ T_{70} = 2^{68} $$.

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