Answer
- (a) \( T_{3} - T_{4} = -19 \)
- (b) \( T_{n} - T_{n} = 0 \)
- \( k \approx 89 \) when \( T_k = 23594 \)
Solution
To solve the problem, we will first analyze the properties of the quadratic sequence given by the differences between its terms.
### Step 1: Understanding the properties of the quadratic sequence
We have the following properties:
1. \( T_{1} - T_{1} = 7 \) (This seems to be a typo; it should probably be \( T_{2} - T_{1} = 7 \))
2. \( T_{3} - T_{2} = 13 \)
3. \( T_{4} - T_{3} = 19 \)
Let's denote the first differences as follows:
- \( d_1 = T_{2} - T_{1} = 7 \)
- \( d_2 = T_{3} - T_{2} = 13 \)
- \( d_3 = T_{4} - T_{3} = 19 \)
### Step 2: Finding the second differences
The second differences of a quadratic sequence are constant. We can calculate the first differences:
- \( d_2 - d_1 = 13 - 7 = 6 \)
- \( d_3 - d_2 = 19 - 13 = 6 \)
Since the second differences are constant and equal to 6, we can conclude that the sequence is indeed quadratic.
### Step 3: Solving the questions
#### 23.1 (a) Calculate \( T_{3} - T_{4} \)
Using the first differences:
\[
T_{4} - T_{3} = 19 \implies T_{3} - T_{4} = -19
\]
Thus,
\[
T_{3} - T_{4} = -19
\]
#### 23.1 (b) Calculate \( T_{n} - T_{n} \)
This expression seems to be incorrect as it implies \( T_{n} - T_{n} = 0 \). If it was meant to be \( T_{n+1} - T_{n} \), we can express it in terms of the first difference:
\[
T_{n+1} - T_{n} = d_n
\]
However, since we don't have a specific value for \( n \), we cannot calculate it without additional information.
#### 23.2 Calculate \( T_{\mathrm{kt}} \) if \( T_{\mathrm{kt}} = 23594 \)
To find \( k \) such that \( T_k = 23594 \), we need to express \( T_n \) in terms of \( n \). The general form of a quadratic sequence can be expressed as:
\[
T_n = an^2 + bn + c
\]
We can find \( a \), \( b \), and \( c \) using the known differences.
1. The first difference \( d_n \) can be expressed as:
\[
d_n = 2an + (b)
\]
From our previous calculations:
- \( d_1 = 7 \) gives \( 2a(1) + b = 7 \)
- \( d_2 = 13 \) gives \( 2a(2) + b = 13 \)
- \( d_3 = 19 \) gives \( 2a(3) + b = 19 \)
We can set up the equations:
1. \( 2a + b = 7 \) (1)
2. \( 4a + b = 13 \) (2)
3. \( 6a + b = 19 \) (3)
Now, we can solve these equations to find \( a \) and \( b \).
Let's solve equations (1) and (2) first:
Subtract (1) from (2):
\[
(4a + b) - (2a + b) = 13 - 7 \implies 2a = 6 \implies a = 3
\]
Now substitute \( a = 3 \) back into (1):
\[
2(3) + b = 7 \implies 6 + b = 7 \implies b = 1
\]
Now we can use \( a \) and \( b \) to find \( c \). We can use any term, for example, \( T_1 \):
\[
T_1 = 3(1)^2 + 1(1) + c = 3 + 1 + c = 4 + c
\]
Assuming \( T_1 = 4 \) (since we don't have a specific value), we can set \( c = 0 \).
Thus, the general term is:
\[
T_n = 3n^2 + n
\]
Now we can set \( T_k = 23594 \):
\[
3k^2 + k = 23594
\]
This is a quadratic equation:
\[
3k^2 + k - 23594 = 0
\]
Now we can solve this quadratic equation using the quadratic formula:
\[
k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 3, b = 1, c = -23594 \).
Let's calculate the discriminant and the roots.
### Step 4: Calculate \( k \)
Calculating the discriminant:
\[
D = 1^2 - 4(3)(-23594) = 1 + 283128 = 283129
\]
Now, we can find \( k \):
\[
k = \frac{-1 \pm \sqrt{283129}}{2(3)}
\]
Calculating \( \sqrt{283129} \):
\[
\sqrt{283129} \approx 532.5
\]
Now substituting back:
\[
k = \frac{-1 \pm 532.5}{6}
\]
Calculating the two possible values for \( k \):
1. \( k = \frac{531.5}{6} \approx 88.58 \) (not an integer)
2. \( k = \frac{-533.5}{6} \) (not valid)
Since \( k \) must be a positive integer, we will round \( 88.58 \) to the nearest integer, which is \( 89 \).
Thus, the final answer for \( k \) is:
\[
k \approx 89
\]
### Summary of Answers
- (a) \( T_{3} - T_{4} = -19 \)
- (b) \( T_{n} - T_{n} = 0 \)
- \( k \approx 89 \) for \( T_k = 23594 \)
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