At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Let's tackle the problem step-by-step for both the arithmetic progression (AP) and the geometric progression (GP).
### Arithmetic Progression (AP) Case
1. Identify the terms:
- First term ([tex]\(a_1\)[/tex]) = [tex]\(-7x\)[/tex]
- Second term ([tex]\(a_2\)[/tex]) = [tex]\(x^2\)[/tex]
2. Common difference ([tex]\(d\)[/tex]):
- Given: [tex]\(d = -6\)[/tex]
3. Using the formula for the nth term of an arithmetic progression:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
For the second term ([tex]\(a_2\)[/tex]):
[tex]\[ a_2 = a_1 + d \][/tex]
Substitute the given values:
[tex]\[ x^2 = -7x + (-6) \][/tex]
4. Solve for [tex]\(x\)[/tex]:
[tex]\[ x^2 = -7x - 6 \][/tex]
Rearrange the equation to standard quadratic form:
[tex]\[ x^2 + 7x + 6 = 0 \][/tex]
Factorize the quadratic equation:
[tex]\[ (x + 6)(x + 1) = 0 \][/tex]
Thus, the possible values of [tex]\(x\)[/tex] are:
[tex]\[ x = -6 \quad \text{and} \quad x = -1 \][/tex]
5. Find the fifth term ([tex]\(a_5\)[/tex]):
Using the nth term formula for [tex]\(a_5\)[/tex]:
[tex]\[ a_5 = a_1 + 4d \][/tex]
Substitute [tex]\(a_1\)[/tex] and [tex]\(d\)[/tex]:
For [tex]\(x = -6\)[/tex]:
[tex]\[ a_1 = -7(-6) = 42 \][/tex]
[tex]\[ a_5 = 42 + 4(-6) = 42 - 24 = 18 \][/tex]
For [tex]\(x = -1\)[/tex]:
[tex]\[ a_1 = -7(-1) = 7 \][/tex]
[tex]\[ a_5 = 7 + 4(-6) = 7 - 24 = -17 \][/tex]
Therefore, the possible values of [tex]\(x\)[/tex] are [tex]\(-6\)[/tex] and [tex]\(-1\)[/tex], with the corresponding fifth terms being [tex]\(18\)[/tex] and [tex]\(-17\)[/tex], respectively.
### Geometric Progression (GP) Case
1. Identify the terms:
- First term ([tex]\(g_1\)[/tex]) = [tex]\(-7x\)[/tex]
- Second term ([tex]\(g_2\)[/tex]) = [tex]\(x^2\)[/tex]
2. Sum to infinity ([tex]\(S_{\infty}\)[/tex]):
- Given: [tex]\(S_{\infty} = -14\)[/tex]
- The sum to infinity formula for a geometric series is:
[tex]\[ S_{\infty} = \frac{g_1}{1 - r} \][/tex]
Where [tex]\(r\)[/tex] is the common ratio.
3. Using the sum to infinity formula:
[tex]\[ -14 = \frac{-7x}{1 - r} \][/tex]
Simplify to find [tex]\(r\)[/tex]:
[tex]\[ -14(1 - r) = -7x \][/tex]
[tex]\[ 14(1 - r) = 7x \][/tex]
[tex]\[ 2(1 - r) = x \][/tex]
[tex]\[ x = 2 - 2r \][/tex]
4. Relate the first term and second term using the common ratio:
[tex]\[ g_2 = g_1 \cdot r \][/tex]
[tex]\[ x^2 = (-7x)r \][/tex]
Substitute [tex]\(x\)[/tex] with the values found from the AP case. Let’s use [tex]\(x = -6\)[/tex]:
For [tex]\(x = -6\)[/tex]:
[tex]\[ (-6)^2 = (-7(-6))r \][/tex]
[tex]\[ 36 = 42r \][/tex]
[tex]\[ r = \frac{36}{42} = \frac{6}{7} \][/tex]
5. Find the third term ([tex]\(g_3\)[/tex]):
[tex]\[ g_3 = g_2 \cdot r \][/tex]
[tex]\[ g_2 = (-7x)r = (-7(-6))\left(\frac{6}{7}\right) = 42 \cdot \frac{6}{7} = 36 \][/tex]
[tex]\[ g_3 = 36 \left(\frac{6}{7}\right) = \frac{216}{7} \][/tex]
Hence, for the geometric progression with [tex]\(x = -6\)[/tex], the third term is [tex]\(\frac{216}{7}\)[/tex] and the common ratio [tex]\(r\)[/tex] is [tex]\(\frac{6}{7}\)[/tex].
### Arithmetic Progression (AP) Case
1. Identify the terms:
- First term ([tex]\(a_1\)[/tex]) = [tex]\(-7x\)[/tex]
- Second term ([tex]\(a_2\)[/tex]) = [tex]\(x^2\)[/tex]
2. Common difference ([tex]\(d\)[/tex]):
- Given: [tex]\(d = -6\)[/tex]
3. Using the formula for the nth term of an arithmetic progression:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
For the second term ([tex]\(a_2\)[/tex]):
[tex]\[ a_2 = a_1 + d \][/tex]
Substitute the given values:
[tex]\[ x^2 = -7x + (-6) \][/tex]
4. Solve for [tex]\(x\)[/tex]:
[tex]\[ x^2 = -7x - 6 \][/tex]
Rearrange the equation to standard quadratic form:
[tex]\[ x^2 + 7x + 6 = 0 \][/tex]
Factorize the quadratic equation:
[tex]\[ (x + 6)(x + 1) = 0 \][/tex]
Thus, the possible values of [tex]\(x\)[/tex] are:
[tex]\[ x = -6 \quad \text{and} \quad x = -1 \][/tex]
5. Find the fifth term ([tex]\(a_5\)[/tex]):
Using the nth term formula for [tex]\(a_5\)[/tex]:
[tex]\[ a_5 = a_1 + 4d \][/tex]
Substitute [tex]\(a_1\)[/tex] and [tex]\(d\)[/tex]:
For [tex]\(x = -6\)[/tex]:
[tex]\[ a_1 = -7(-6) = 42 \][/tex]
[tex]\[ a_5 = 42 + 4(-6) = 42 - 24 = 18 \][/tex]
For [tex]\(x = -1\)[/tex]:
[tex]\[ a_1 = -7(-1) = 7 \][/tex]
[tex]\[ a_5 = 7 + 4(-6) = 7 - 24 = -17 \][/tex]
Therefore, the possible values of [tex]\(x\)[/tex] are [tex]\(-6\)[/tex] and [tex]\(-1\)[/tex], with the corresponding fifth terms being [tex]\(18\)[/tex] and [tex]\(-17\)[/tex], respectively.
### Geometric Progression (GP) Case
1. Identify the terms:
- First term ([tex]\(g_1\)[/tex]) = [tex]\(-7x\)[/tex]
- Second term ([tex]\(g_2\)[/tex]) = [tex]\(x^2\)[/tex]
2. Sum to infinity ([tex]\(S_{\infty}\)[/tex]):
- Given: [tex]\(S_{\infty} = -14\)[/tex]
- The sum to infinity formula for a geometric series is:
[tex]\[ S_{\infty} = \frac{g_1}{1 - r} \][/tex]
Where [tex]\(r\)[/tex] is the common ratio.
3. Using the sum to infinity formula:
[tex]\[ -14 = \frac{-7x}{1 - r} \][/tex]
Simplify to find [tex]\(r\)[/tex]:
[tex]\[ -14(1 - r) = -7x \][/tex]
[tex]\[ 14(1 - r) = 7x \][/tex]
[tex]\[ 2(1 - r) = x \][/tex]
[tex]\[ x = 2 - 2r \][/tex]
4. Relate the first term and second term using the common ratio:
[tex]\[ g_2 = g_1 \cdot r \][/tex]
[tex]\[ x^2 = (-7x)r \][/tex]
Substitute [tex]\(x\)[/tex] with the values found from the AP case. Let’s use [tex]\(x = -6\)[/tex]:
For [tex]\(x = -6\)[/tex]:
[tex]\[ (-6)^2 = (-7(-6))r \][/tex]
[tex]\[ 36 = 42r \][/tex]
[tex]\[ r = \frac{36}{42} = \frac{6}{7} \][/tex]
5. Find the third term ([tex]\(g_3\)[/tex]):
[tex]\[ g_3 = g_2 \cdot r \][/tex]
[tex]\[ g_2 = (-7x)r = (-7(-6))\left(\frac{6}{7}\right) = 42 \cdot \frac{6}{7} = 36 \][/tex]
[tex]\[ g_3 = 36 \left(\frac{6}{7}\right) = \frac{216}{7} \][/tex]
Hence, for the geometric progression with [tex]\(x = -6\)[/tex], the third term is [tex]\(\frac{216}{7}\)[/tex] and the common ratio [tex]\(r\)[/tex] is [tex]\(\frac{6}{7}\)[/tex].
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
