Answer :
Let's break down the problem step-by-step:
### Question 8.1: The first three terms of an arithmetic sequence are [tex]\(2p + 3\)[/tex]; [tex]\(p + 6\)[/tex]; and [tex]\(p - 2\)[/tex].
#### Part 8.1.1: Show that [tex]\(p = 11\)[/tex]
In an arithmetic sequence, the difference between consecutive terms is constant. Let's denote this common difference by [tex]\(d\)[/tex].
1. Identify the common difference between the first and second terms:
[tex]\[ d = (p + 6) - (2p + 3) \][/tex]
Simplify this expression:
[tex]\[ d = p + 6 - 2p - 3 = -p + 3 \][/tex]
2. Identify the common difference between the second and third terms:
[tex]\[ d = (p - 2) - (p + 6) \][/tex]
Simplify this expression:
[tex]\[ d = p - 2 - p - 6 = -8 \][/tex]
Since both differences must be the same (because they are consecutive terms in an arithmetic sequence):
[tex]\[ -p + 3 = -8 \][/tex]
Solve for [tex]\(p\)[/tex]:
[tex]\[ -p + 3 = -8 \implies -p = -8 - 3 \implies -p = -11 \implies p = 11 \][/tex]
Therefore, we have shown that [tex]\(p = 11\)[/tex].
#### Part 8.1.2: Calculate the smallest value of [tex]\(n\)[/tex] for which [tex]\(T_n < -55\)[/tex]
The general term [tex]\(T_n\)[/tex] of an arithmetic sequence can be formulated as:
[tex]\[ T_n = a + (n - 1)d \][/tex]
where [tex]\(a\)[/tex] is the first term and [tex]\(d\)[/tex] is the common difference.
From part 8.1.1, we know:
- [tex]\(p = 11\)[/tex]
- The terms are: [tex]\(2p + 3\)[/tex], [tex]\(p + 6\)[/tex], and [tex]\(p - 2\)[/tex]:
[tex]\[ 2(11) + 3 = 22 + 3 = 25 \quad (\text{first term}) \][/tex]
- The common difference [tex]\(d\)[/tex] is:
[tex]\[ (11 + 6) - (2 \times 11 + 3) = -8 \text{ or } (11 - 2) - (11 + 6) = -8 \][/tex]
Now, we need [tex]\(T_n\)[/tex] such that:
[tex]\[ T_n < -55 \][/tex]
Substitute the values:
[tex]\[ 25 + (n - 1)(-8) < -55 \][/tex]
Simplify and solve for [tex]\(n\)[/tex]:
[tex]\[ 25 - 8(n - 1) < -55 \][/tex]
[tex]\[ 25 - 8n + 8 < -55 \][/tex]
[tex]\[ 33 - 8n < -55 \][/tex]
[tex]\[ -8n < -55 - 33 \][/tex]
[tex]\[ -8n < -88 \][/tex]
[tex]\[ n > \frac{88}{8} \][/tex]
[tex]\[ n > 11 \][/tex]
Thus, the smallest value of [tex]\(n\)[/tex] is [tex]\(n = 12\)[/tex].
### Question 8.2: Given that [tex]\( \sum_{k=1}^6(x-3k)=\sum_{k=1}^9(x-3k)\)[/tex], prove that [tex]\(\sum_{k=1}^{15}(x-3k)=0\)[/tex]
1. Simplify the given sums:
[tex]\[ \sum_{k=1}^6(x - 3k) = \sum_{k=1}^9(x - 3k) \][/tex]
2. Calculate the individual sums:
For [tex]\(\sum_{k=1}^6(x - 3k)\)[/tex]:
[tex]\[ \sum_{k=1}^6(x - 3k) = \sum_{k=1}^6 x - \sum_{k=1}^6 3k = 6x - 3\sum_{k=1}^6 k = 6x - 3 \left( \frac{6(6 + 1)}{2} \right) = 6x - 63 \][/tex]
For [tex]\(\sum_{k=1}^9(x - 3k)\)[/tex]:
[tex]\[ \sum_{k=1}^9(x - 3k) = \sum_{k=1}^9 x - \sum_{k=1}^9 3k = 9x - 3\sum_{k=1}^9 k = 9x - 3 \left( \frac{9(9 + 1)}{2} \right) = 9x - 135 \][/tex]
Given equality:
[tex]\[ 6x - 63 = 9x - 135 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ 6x - 63 = 9x - 135 \implies -63 + 135 = 9x - 6x \implies 72 = 3x \implies x = 24 \][/tex]
3. Sum for [tex]\(k = 1\)[/tex] to [tex]\(15\)[/tex]:
[tex]\[ \sum_{k=1}^{15} (x - 3k) = \sum_{k=1}^{15} 24 - 3\sum_{k=1}^{15} k = 15(24) - 3 \left( \frac{15(15 + 1)}{2} \right) \][/tex]
[tex]\[ = 360 - 3(120) = 360 - 360 = 0 \][/tex]
Thus, we have proved that:
[tex]\[ \sum_{k=1}^{15}(x - 3k) = 0 \][/tex]
### Question 8.1: The first three terms of an arithmetic sequence are [tex]\(2p + 3\)[/tex]; [tex]\(p + 6\)[/tex]; and [tex]\(p - 2\)[/tex].
#### Part 8.1.1: Show that [tex]\(p = 11\)[/tex]
In an arithmetic sequence, the difference between consecutive terms is constant. Let's denote this common difference by [tex]\(d\)[/tex].
1. Identify the common difference between the first and second terms:
[tex]\[ d = (p + 6) - (2p + 3) \][/tex]
Simplify this expression:
[tex]\[ d = p + 6 - 2p - 3 = -p + 3 \][/tex]
2. Identify the common difference between the second and third terms:
[tex]\[ d = (p - 2) - (p + 6) \][/tex]
Simplify this expression:
[tex]\[ d = p - 2 - p - 6 = -8 \][/tex]
Since both differences must be the same (because they are consecutive terms in an arithmetic sequence):
[tex]\[ -p + 3 = -8 \][/tex]
Solve for [tex]\(p\)[/tex]:
[tex]\[ -p + 3 = -8 \implies -p = -8 - 3 \implies -p = -11 \implies p = 11 \][/tex]
Therefore, we have shown that [tex]\(p = 11\)[/tex].
#### Part 8.1.2: Calculate the smallest value of [tex]\(n\)[/tex] for which [tex]\(T_n < -55\)[/tex]
The general term [tex]\(T_n\)[/tex] of an arithmetic sequence can be formulated as:
[tex]\[ T_n = a + (n - 1)d \][/tex]
where [tex]\(a\)[/tex] is the first term and [tex]\(d\)[/tex] is the common difference.
From part 8.1.1, we know:
- [tex]\(p = 11\)[/tex]
- The terms are: [tex]\(2p + 3\)[/tex], [tex]\(p + 6\)[/tex], and [tex]\(p - 2\)[/tex]:
[tex]\[ 2(11) + 3 = 22 + 3 = 25 \quad (\text{first term}) \][/tex]
- The common difference [tex]\(d\)[/tex] is:
[tex]\[ (11 + 6) - (2 \times 11 + 3) = -8 \text{ or } (11 - 2) - (11 + 6) = -8 \][/tex]
Now, we need [tex]\(T_n\)[/tex] such that:
[tex]\[ T_n < -55 \][/tex]
Substitute the values:
[tex]\[ 25 + (n - 1)(-8) < -55 \][/tex]
Simplify and solve for [tex]\(n\)[/tex]:
[tex]\[ 25 - 8(n - 1) < -55 \][/tex]
[tex]\[ 25 - 8n + 8 < -55 \][/tex]
[tex]\[ 33 - 8n < -55 \][/tex]
[tex]\[ -8n < -55 - 33 \][/tex]
[tex]\[ -8n < -88 \][/tex]
[tex]\[ n > \frac{88}{8} \][/tex]
[tex]\[ n > 11 \][/tex]
Thus, the smallest value of [tex]\(n\)[/tex] is [tex]\(n = 12\)[/tex].
### Question 8.2: Given that [tex]\( \sum_{k=1}^6(x-3k)=\sum_{k=1}^9(x-3k)\)[/tex], prove that [tex]\(\sum_{k=1}^{15}(x-3k)=0\)[/tex]
1. Simplify the given sums:
[tex]\[ \sum_{k=1}^6(x - 3k) = \sum_{k=1}^9(x - 3k) \][/tex]
2. Calculate the individual sums:
For [tex]\(\sum_{k=1}^6(x - 3k)\)[/tex]:
[tex]\[ \sum_{k=1}^6(x - 3k) = \sum_{k=1}^6 x - \sum_{k=1}^6 3k = 6x - 3\sum_{k=1}^6 k = 6x - 3 \left( \frac{6(6 + 1)}{2} \right) = 6x - 63 \][/tex]
For [tex]\(\sum_{k=1}^9(x - 3k)\)[/tex]:
[tex]\[ \sum_{k=1}^9(x - 3k) = \sum_{k=1}^9 x - \sum_{k=1}^9 3k = 9x - 3\sum_{k=1}^9 k = 9x - 3 \left( \frac{9(9 + 1)}{2} \right) = 9x - 135 \][/tex]
Given equality:
[tex]\[ 6x - 63 = 9x - 135 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ 6x - 63 = 9x - 135 \implies -63 + 135 = 9x - 6x \implies 72 = 3x \implies x = 24 \][/tex]
3. Sum for [tex]\(k = 1\)[/tex] to [tex]\(15\)[/tex]:
[tex]\[ \sum_{k=1}^{15} (x - 3k) = \sum_{k=1}^{15} 24 - 3\sum_{k=1}^{15} k = 15(24) - 3 \left( \frac{15(15 + 1)}{2} \right) \][/tex]
[tex]\[ = 360 - 3(120) = 360 - 360 = 0 \][/tex]
Thus, we have proved that:
[tex]\[ \sum_{k=1}^{15}(x - 3k) = 0 \][/tex]
