Discover a wealth of information and get your questions answered on IDNLearn.com. Discover in-depth answers to your questions from our community of experienced professionals.
Sagot :
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]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For precise answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
