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[Maximum mark: 18] The first three terms of an infinite sequence, in order, are...
Jul 20, 2024
Solution by Steps
step 1
Consider the given geometric series: 2ln⁡x,qln⁡x,ln⁡x2 \ln x, q \ln x, \ln \sqrt{x}
step 2
For the series to be geometric, the ratio between consecutive terms must be constant. Thus, qln⁡x2ln⁡x=ln⁡xqln⁡x\frac{q \ln x}{2 \ln x} = \frac{\ln \sqrt{x}}{q \ln x}
step 3
Simplify the ratios: q2=ln⁡xqln⁡x\frac{q}{2} = \frac{\ln \sqrt{x}}{q \ln x}
step 4
Note that ln⁡x=12ln⁡x\ln \sqrt{x} = \frac{1}{2} \ln x, so q2=12ln⁡xqln⁡x\frac{q}{2} = \frac{\frac{1}{2} \ln x}{q \ln x}
step 5
Simplify further: q2=12q\frac{q}{2} = \frac{1}{2q}
step 6
Solve for qq: q2=1  ⟹  q=±1q^2 = 1 \implies q = \pm 1
step 7
Since q > 0, we have q=1q = 1
Answer
q=1q = 1
Key Concept
Geometric series ratio
Explanation
For a series to be geometric, the ratio between consecutive terms must be constant.
Solution by Steps
step 1
To show the series is convergent, we need to check if the common ratio |r| < 1
step 2
Since q=1q = 1, the common ratio r=qln⁡x2ln⁡x=12r = \frac{q \ln x}{2 \ln x} = \frac{1}{2}
step 3
Since |r| = \frac{1}{2} < 1, the series is convergent
Answer
The series is convergent because |r| = \frac{1}{2} < 1
Key Concept
Convergence of geometric series
Explanation
A geometric series converges if the absolute value of the common ratio is less than 1.
Solution by Steps
step 1
Given q > 0 and S∞=8ln⁡3S_{\infty} = 8 \ln 3, we use the sum formula for an infinite geometric series: S∞=a1−rS_{\infty} = \frac{a}{1 - r}
step 2
Here, a=2ln⁡xa = 2 \ln x and r=12r = \frac{1}{2}
step 3
Substitute the values: 8ln⁡3=2ln⁡x1−128 \ln 3 = \frac{2 \ln x}{1 - \frac{1}{2}}
step 4
Simplify: 8ln⁡3=4ln⁡x8 \ln 3 = 4 \ln x
step 5
Solve for xx: ln⁡x=2ln⁡3  ⟹  x=32=9\ln x = 2 \ln 3 \implies x = 3^2 = 9
Answer
x=9x = 9
Key Concept
Sum of an infinite geometric series
Explanation
The sum of an infinite geometric series is given by a1−r\frac{a}{1 - r}, where aa is the first term and rr is the common ratio.
Solution by Steps
step 1
For the series to be arithmetic, the difference between consecutive terms must be constant
step 2
Given terms: 2ln⁡x,qln⁡x,ln⁡x2 \ln x, q \ln x, \ln \sqrt{x}
step 3
The common difference d=qln⁡x−2ln⁡x=ln⁡x−qln⁡xd = q \ln x - 2 \ln x = \ln \sqrt{x} - q \ln x
step 4
Simplify: d=(q−2)ln⁡x=12ln⁡x−qln⁡xd = (q - 2) \ln x = \frac{1}{2} \ln x - q \ln x
step 5
Equate the differences: q−2=12−qq - 2 = \frac{1}{2} - q
step 6
Solve for qq: 2q−2=12  ⟹  2q=52  ⟹  q=542q - 2 = \frac{1}{2} \implies 2q = \frac{5}{2} \implies q = \frac{5}{4}
Answer
q=54q = \frac{5}{4}
Key Concept
Arithmetic series common difference
Explanation
For a series to be arithmetic, the difference between consecutive terms must be constant.
Solution by Steps
step 1
The common difference d=qln⁡x−2ln⁡xd = q \ln x - 2 \ln x
step 2
Substitute q=54q = \frac{5}{4}: d=54ln⁡x−2ln⁡xd = \frac{5}{4} \ln x - 2 \ln x
step 3
Simplify: d=(54−2)ln⁡x=−34ln⁡xd = \left(\frac{5}{4} - 2\right) \ln x = -\frac{3}{4} \ln x
Answer
d=−34ln⁡xd = -\frac{3}{4} \ln x
Key Concept
Common difference in arithmetic series
Explanation
The common difference in an arithmetic series is the difference between consecutive terms.
Solution by Steps
step 1
Given the sum of the first nn terms Sn=ln⁡x5S_n = \ln \sqrt{x^5}
step 2
The sum of the first nn terms of an arithmetic series is Sn=n2(2a+(n−1)d)S_n = \frac{n}{2} (2a + (n-1)d)
step 3
Here, a=2ln⁡xa = 2 \ln x and d=−34ln⁡xd = -\frac{3}{4} \ln x
step 4
Substitute the values: ln⁡x5=n2(2ln⁡x+(n−1)(−34ln⁡x))\ln \sqrt{x^5} = \frac{n}{2} (2 \ln x + (n-1)(-\frac{3}{4} \ln x))
step 5
Simplify: ln⁡x5/2=n2(2ln⁡x−34(n−1)ln⁡x)\ln x^{5/2} = \frac{n}{2} (2 \ln x - \frac{3}{4} (n-1) \ln x)
step 6
Further simplify: 52ln⁡x=n2(2−34(n−1))ln⁡x\frac{5}{2} \ln x = \frac{n}{2} \left(2 - \frac{3}{4} (n-1)\right) \ln x
step 7
Cancel ln⁡x\ln x: 52=n2(2−34(n−1))\frac{5}{2} = \frac{n}{2} \left(2 - \frac{3}{4} (n-1)\right)
step 8
Solve for nn: 5=n(2−34(n−1))5 = n \left(2 - \frac{3}{4} (n-1)\right)
step 9
Simplify: 5=2n−34n2+34n5 = 2n - \frac{3}{4} n^2 + \frac{3}{4} n
step 10
Combine like terms: 5=2n+34n−34n25 = 2n + \frac{3}{4} n - \frac{3}{4} n^2
step 11
Further simplify: 5=114n−34n25 = \frac{11}{4} n - \frac{3}{4} n^2
step 12
Multiply by 4: 20=11n−3n220 = 11n - 3n^2
step 13
Rearrange: 3n2−11n+20=03n^2 - 11n + 20 = 0
step 14
Solve the quadratic equation: n=11±121−2406n = \frac{11 \pm \sqrt{121 - 240}}{6}
step 15
Since the discriminant is negative, there is no real solution for nn
Answer
There is no real value for nn that satisfies the given conditions.
Key Concept
Sum of arithmetic series
Explanation
The sum of the first nn terms of an arithmetic series is given by n2(2a+(n−1)d)\frac{n}{2} (2a + (n-1)d).
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