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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].
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