Question
QUESTION 23 (NOVEMBER 2014) 23.1 Given the quadratic sequence: \( -1 ;-7 ;-11 ; p ; \ldots \) 23.1.1 Write down the value of \( p \). (2) 23.1.2 Determine the \( n^{\text {th }} \) term of the sequence. (4) 23.1.3 The first difference between consecutive terms of the sequence is 96 calculate the values of these two terms (4) 23.2 The first three terms of a geometric sequence are: \( 16 ; 4 ; 1 \) 23.2.1 Calculate the value of the \( 12^{\text {th }} \) term. (Leave your answer in simplified exponential form.) 23.2.2 Calculate the sum of the first 10 terms of the sequence. (2) 23.3 Determine the value of: \( \quad\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{5}\right) \ldots \) up to 98 factors
Ask by Reed Wang. in South Africa
Jan 26,2025
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Tutor-Verified Answer
Answer
**23.1 Quadratic Sequence**
1. **Value of \( p \):**
- The sequence is \( -1, -7, -11, p, \ldots \).
- First differences: \( -6, -4 \).
- Second difference: \( 2 \).
- Next first difference: \( -2 \).
- \( p = -11 + (-2) = -13 \).
- **Answer:** \( p = -13 \).
2. **\( n^{\text{th}} \) Term:**
- Using the quadratic formula: \( a_n = n^2 - 9n + 7 \).
- **Answer:** \( a_n = n^2 - 9n + 7 \).
3. **Terms with First Difference 96:**
- \( a_{52} = 2243 \) and \( a_{53} = 2339 \).
- **Answer:** \( 2243 \) and \( 2339 \).
**23.2 Geometric Sequence**
1. **12th Term:**
- Common ratio \( r = \frac{1}{4} \).
- \( a_{12} = 16 \times \left(\frac{1}{4}\right)^{11} = 2^{-18} \).
- **Answer:** \( 2^{-18} \).
2. **Sum of First 10 Terms:**
- Sum \( S_{10} = 16 \times \frac{1 - \left(\frac{1}{4}\right)^{10}}{1 - \frac{1}{4}} = 16 \times \frac{1 - \frac{1}{4^{10}}}{\frac{3}{4}} = \frac{64}{3} \left(1 - \frac{1}{4^{10}}\right) \).
- **Answer:** \( \frac{64}{3} \left(1 - \frac{1}{4^{10}}\right) \).
**23.3 Product of Series**
- Calculate the product: \( \left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right)\left(1 + \frac{1}{4}\right)\left(1 + \frac{1}{5}\right) \ldots \) up to 98 factors.
- This is a telescoping product where most terms cancel out, resulting in \( \frac{99}{2} \).
- **Answer:** \( \frac{99}{2} \).
Solution
Let's solve the problem step by step.
### 23.1 Quadratic Sequence
#### 23.1.1 Write down the value of \( p \).
The given sequence is \( -1, -7, -11, p, \ldots \).
To find \( p \), we first need to determine the pattern in the sequence. We can calculate the first differences between consecutive terms:
- First difference between \( -1 \) and \( -7 \):
\[
-7 - (-1) = -6
\]
- First difference between \( -7 \) and \( -11 \):
\[
-11 - (-7) = -4
\]
The first differences are \( -6 \) and \( -4 \). The second difference (which should be constant for a quadratic sequence) is:
\[
-4 - (-6) = 2
\]
Now, we can find the next first difference by adding \( 2 \) to the last first difference:
\[
-4 + 2 = -2
\]
Now, we can find \( p \) using the last first difference:
\[
p - (-11) = -2 \implies p + 11 = -2 \implies p = -2 - 11 = -13
\]
Thus, the value of \( p \) is:
\[
\boxed{-13}
\]
#### 23.1.2 Determine the \( n^{\text{th}} \) term of the sequence.
The first differences we calculated are \( -6, -4, -2 \). The second differences are constant at \( 2 \).
The first term \( a_1 = -1 \), and the first difference sequence is \( -6, -4, -2 \).
The \( n^{\text{th}} \) term of a quadratic sequence can be expressed as:
\[
a_n = An^2 + Bn + C
\]
We know:
- \( a_1 = A(1)^2 + B(1) + C = -1 \)
- \( a_2 = A(2)^2 + B(2) + C = -7 \)
- \( a_3 = A(3)^2 + B(3) + C = -11 \)
This gives us the following system of equations:
1. \( A + B + C = -1 \)
2. \( 4A + 2B + C = -7 \)
3. \( 9A + 3B + C = -11 \)
We can solve this system of equations. First, we can eliminate \( C \) by subtracting the first equation from the second and the second from the third.
Subtracting equation 1 from equation 2:
\[
(4A + 2B + C) - (A + B + C) = -7 - (-1) \implies 3A + B = -6 \quad \text{(Equation 4)}
\]
Subtracting equation 2 from equation 3:
\[
(9A + 3B + C) - (4A + 2B + C) = -11 - (-7) \implies 5A + B = -4 \quad \text{(Equation 5)}
\]
Now, we can subtract equation 4 from equation 5:
\[
(5A + B) - (3A + B) = -4 - (-6) \implies 2A = 2 \implies A = 1
\]
Substituting \( A = 1 \) back into equation 4:
\[
3(1) + B = -6 \implies 3 + B = -6 \implies B = -9
\]
Now substituting \( A \) and \( B \) back into equation 1 to find \( C \):
\[
1 - 9 + C = -1 \implies C = 7
\]
Thus, the \( n^{\text{th}} \) term is:
\[
a_n = n^2 - 9n + 7
\]
So, the \( n^{\text{th}} \) term of the sequence is:
\[
\boxed{n^2 - 9n + 7}
\]
#### 23.1.3 The first difference between consecutive terms of the sequence is 96. Calculate the values of these two terms.
The first difference is given as \( 96 \). The first difference between two consecutive terms \( a_n \) and \( a_{n+1} \) is:
\[
a_{n+1} - a_n = 96
\]
Using the formula for the \( n^{\text{th}} \) term:
\[
(n+1)^2 - 9(n+1) + 7 - (n^2 - 9n + 7) = 96
\]
Simplifying this:
\[
(n^2 + 2n + 1 - 9n - 9 + 7) - (n^2 - 9n + 7) = 96
\]
\[
(2n + 1 - 9) = 96
\]
\[
2n - 8 = 96 \implies 2n = 104 \implies n = 52
\]
Now, we can find the two terms:
\[
a_{52} = 52^2 - 9(52) + 7
\]
\[
= 2704 - 468 + 7 = 2243
\]
\[
a_{53} = 53^2 - 9(53) + 7
\]
\[
= 2809 - 477 + 7 = 2339
\]
Thus, the two terms are:
\[
\boxed{2243 \text{ and } 2339}
\]
### 23.2 Geometric Sequence
#### 23.2.1 Calculate the value of the \( 12^{\text{th}} \) term.
The first three terms of the geometric sequence are \( 16, 4, 1 \).
To find the common ratio \( r \):
\[
r = \frac{4}{16} = \frac{1}{4}
\]
The \( n^{\text{th}} \) term of a geometric sequence is given by:
\[
a_n = a_1 \cdot r^{n-1}
\]
For the \( 12^{\text{th}} \) term:
\[
a_{12} = 16 \cdot \left(\frac{1}{4}\right)^{11}
\]
\[
= 16 \cdot \frac{1}{4^{11}} = 16 \cdot \frac{1}{2^{22}} = \frac{16}{2^{22}} = \frac{2^4}{2^{22}} = \frac{1}{2^{18}} = 2^{-18}
\]
Thus, the \(
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Simplify this solution The Deep Dive
In the quadratic sequence \( -1, -7, -11, p, \ldots \), the differences between terms are \( -6 \) and \( -4 \). To find \( p \), we realize the next difference will be \( -2 \). So, \( p = -11 - 2 = -13 \). The \( n^{\text{th}} \) term of this sequence can be written as \( T_n = -1 + (n-1)(-6 + (n-2)) \), which simplifies to a quadratic expression. This becomes \( T_n = - 2n^2 + 10n - 1 \) after some algebra rearrangement. You can use quadratic sequences in real life with things like architecture or nature, where patterns seem to recur in parabolic forms. For example, the archways in buildings or bridges often mimic the shapes found in quadratic functions. Exploring sequences can lead you down some interesting paths, like mathematical modeling or even programming applications. If sequences interest you, check out resources like "The Art of Problem Solving" by Richard Rusczyk, which delves into the exciting world of sequences and series!
