🧠:<think>Okay, so I have this problem about Kai, a martial artist who logs his training hours and tournament scores over 10 weeks. There are two parts to the problem. Let me tackle them one by one.Starting with the first question: It says that Kai's training hours form an arithmetic progression. The total over 10 weeks is 250 hours, and he trained for 15 hours in the first week. I need to find the common difference of this arithmetic progression.Alright, arithmetic progression. I remember that an arithmetic progression is a sequence where each term after the first is obtained by adding a constant difference. So, if the first term is a1, then the second term is a1 + d, the third is a1 + 2d, and so on. The nth term is a1 + (n-1)d.Given that, the total training hours over 10 weeks would be the sum of the first 10 terms of this arithmetic progression. The formula for the sum of the first n terms of an arithmetic progression is S_n = n/2 * (2a1 + (n-1)d). Alternatively, it can also be written as S_n = n * (a1 + an)/2, where an is the nth term.Let me write down the given values:- Number of weeks, n = 10- Total training hours, S_10 = 250- First week's training hours, a1 = 15I need to find the common difference, d.Using the sum formula:S_10 = 10/2 * (2*15 + (10 - 1)*d) Simplify that:250 = 5 * (30 + 9d) Divide both sides by 5:50 = 30 + 9d Subtract 30 from both sides:20 = 9d So, d = 20 / 9Wait, 20 divided by 9 is approximately 2.222... So, is that the common difference? Hmm, that seems a bit unusual because training hours are usually in whole numbers, but maybe not necessarily. Let me double-check my calculations.Starting again:Sum formula: S_n = n/2 [2a1 + (n - 1)d]Plugging in the numbers:250 = 10/2 [2*15 + 9d] 250 = 5 [30 + 9d] Divide both sides by 5: 50 = 30 + 9d Subtract 30: 20 = 9d So, d = 20/9 ≈ 2.222...Hmm, okay, so that's correct. The common difference is 20/9 hours per week. That's approximately 2.222 hours each week. So, each week he increases his training by about 2.222 hours. That seems a bit precise, but maybe Kai is very precise with his training.Alternatively, maybe I made a mistake in the formula? Let me check.Wait, another version of the sum formula is S_n = n*(a1 + an)/2. Maybe I can use that to verify.First, I need to find a10, the 10th term, which is a1 + 9d. Then, the sum would be 10*(15 + a10)/2 = 250.So, 10*(15 + a10)/2 = 250 Simplify: 5*(15 + a10) = 250 Divide both sides by 5: 15 + a10 = 50 So, a10 = 35Therefore, a10 = 15 + 9d = 35 So, 9d = 35 - 15 = 20 Thus, d = 20/9Same result. So, that's correct. So, the common difference is 20/9 hours per week.Alright, moving on to the second question.Kai's tournament scores form a geometric progression. The first week he scored 100 points, and in the tenth week, he scored 102,400 points. I need to find the common ratio of the geometric progression and determine his score in the 7th week.Okay, geometric progression. Each term is the previous term multiplied by a common ratio, r. So, the nth term is a1 * r^(n-1).Given:- First term, a1 = 100- Tenth term, a10 = 102,400- Number of weeks, n = 10I need to find the common ratio, r, and then find the 7th term, a7.First, let's find r.The formula for the nth term of a geometric progression is:a_n = a1 * r^(n - 1)So, for the 10th term:a10 = 100 * r^(10 - 1) 102,400 = 100 * r^9Divide both sides by 100:1024 = r^9So, r^9 = 1024Hmm, 1024 is a power of 2. Let me recall: 2^10 = 1024, so 1024 is 2^10.But here, r^9 = 2^10So, r = (2^10)^(1/9) = 2^(10/9)Wait, 10/9 is approximately 1.111, so r is 2^(10/9). Let me compute that.But 2^(10/9) is the same as the 9th root of 2^10, which is 2^(1 + 1/9) = 2 * 2^(1/9). Hmm, that's approximately 2 * 1.0801 = 2.1602.But maybe I can express it as 2^(10/9). Alternatively, is 1024 a power of something else?Wait, 1024 is 2^10, 4^5, 8^3.2, but 1024 is 2^10. So, r^9 = 2^10.Therefore, r = 2^(10/9). Alternatively, 2^(1 + 1/9) = 2 * 2^(1/9). So, that's the exact value.Alternatively, maybe 1024 is 4^5, so 4^5 = (2^2)^5 = 2^10. So, r^9 = 4^5. Therefore, r = 4^(5/9). Hmm, that's another way to write it.But perhaps the problem expects an exact value, so 2^(10/9) is the exact common ratio.Alternatively, maybe it's a whole number? Let me check.Wait, 1024 is 2^10, so r^9 = 2^10. So, r = 2^(10/9). That's approximately 2.1602, as I calculated before.Wait, but 2^10 is 1024, so 2^(10/9) is the 9th root of 1024. Let me compute 1024^(1/9). Let me see:Compute 2^10 = 1024.So, 2^10 = (2^(10/9))^9.Therefore, 2^(10/9) is the common ratio.Alternatively, maybe I can write it as 2^(10/9) or 2^(1 + 1/9) = 2 * 2^(1/9). But 2^(1/9) is the 9th root of 2, which is approximately 1.0801. So, 2 * 1.0801 ≈ 2.1602.But perhaps the problem expects an exact value, so 2^(10/9). Alternatively, maybe 1024 is 4^5, so 4^(5/9). Let me compute 4^(5/9):4^(5/9) = (2^2)^(5/9) = 2^(10/9). So, same as before.Alternatively, maybe 1024 is 8^3.2? But 8 is 2^3, so 8^3.2 = (2^3)^3.2 = 2^(9.6). Hmm, not helpful.So, I think the exact value is 2^(10/9). Alternatively, 2^(10/9) can be written as 2^(1 + 1/9) = 2 * 2^(1/9). But 2^(1/9) is irrational, so it's better to leave it as 2^(10/9).Alternatively, maybe the problem expects a fractional exponent or a radical form. 2^(10/9) is equal to the 9th root of 2^10, which is the 9th root of 1024. So, it's √[9]{1024}.But perhaps the problem expects a decimal approximation? Let me see.Compute 2^(10/9):First, 10/9 is approximately 1.1111.So, 2^1 = 22^1.1111 ≈ ?We can use logarithms or natural exponentials.Compute ln(2^1.1111) = 1.1111 * ln(2) ≈ 1.1111 * 0.6931 ≈ 0.7734Then, exponentiate: e^0.7734 ≈ 2.168Wait, that's approximately 2.168.Wait, but earlier I thought it was approximately 2.1602. Hmm, slight discrepancy. Maybe my calculator is better.Alternatively, let me use a calculator:Compute 2^(10/9):10 divided by 9 is approximately 1.1111111111.Compute 2^1.1111111111:We can use the fact that 2^1 = 22^0.1111111111 ≈ ?We know that 2^(1/9) ≈ 1.0801, since 1.0801^9 ≈ 2.Therefore, 2^(10/9) = 2^(1 + 1/9) = 2 * 2^(1/9) ≈ 2 * 1.0801 ≈ 2.1602.So, approximately 2.1602.But let me check with a calculator:Using natural logarithm:ln(2^(10/9)) = (10/9) * ln(2) ≈ (1.1111) * 0.6931 ≈ 0.7734e^0.7734 ≈ e^0.7 * e^0.0734 ≈ 2.0138 * 1.0763 ≈ 2.0138 * 1.0763 ≈ 2.168Hmm, so approximately 2.168. So, maybe 2.168 is a better approximation.But regardless, the exact value is 2^(10/9). So, I think that's the answer for the common ratio.Now, moving on to find the score in the 7th week, which is a7.Using the formula for the nth term of a geometric progression:a7 = a1 * r^(7 - 1) = 100 * r^6We already know that r^9 = 1024, so r^6 = (r^9)^(2/3) = (1024)^(2/3)Compute 1024^(2/3):First, 1024 is 2^10, so 1024^(2/3) = (2^10)^(2/3) = 2^(20/3) = 2^(6 + 2/3) = 2^6 * 2^(2/3) = 64 * (2^(2/3))Now, 2^(2/3) is the cube root of 4, since (2^(2/3))^3 = 2^2 = 4. So, 2^(2/3) ≈ 1.5874Therefore, 64 * 1.5874 ≈ 64 * 1.5874 ≈ 101.44Wait, but let me compute it more accurately.First, 2^(2/3):We know that 2^(1/3) ≈ 1.2599, so 2^(2/3) = (2^(1/3))^2 ≈ (1.2599)^2 ≈ 1.5874So, 64 * 1.5874 ≈ 64 * 1.5874Compute 64 * 1.5 = 9664 * 0.0874 ≈ 64 * 0.08 = 5.12; 64 * 0.0074 ≈ 0.4736So, total ≈ 5.12 + 0.4736 ≈ 5.5936Therefore, total ≈ 96 + 5.5936 ≈ 101.5936So, approximately 101.5936Therefore, a7 = 100 * 101.5936 ≈ 10159.36Wait, that can't be right. Wait, no, hold on.Wait, a7 = 100 * r^6But r^6 is 1024^(2/3) = (2^10)^(2/3) = 2^(20/3) = 2^6 * 2^(2/3) = 64 * 2^(2/3) ≈ 64 * 1.5874 ≈ 101.5936Therefore, a7 = 100 * 101.5936 ≈ 10159.36Wait, but that seems high because in the 10th week, he scored 102,400, which is 100 * r^9 = 100 * 1024 = 102,400. So, in the 7th week, it's 100 * r^6 ≈ 100 * 101.5936 ≈ 10,159.36Wait, but 10,159.36 is much less than 102,400, which is in the 10th week. That seems plausible because the scores are increasing exponentially.Alternatively, maybe I can express a7 in terms of 2^(20/3):a7 = 100 * 2^(20/3) = 100 * 2^(6 + 2/3) = 100 * 64 * 2^(2/3) = 6400 * 2^(2/3)But 2^(2/3) is approximately 1.5874, so 6400 * 1.5874 ≈ 6400 * 1.5874 ≈ 10,159.36So, approximately 10,159.36 points in the 7th week.Alternatively, maybe I can write it as 100 * (2^(10/9))^6 = 100 * 2^(60/9) = 100 * 2^(20/3), which is the same as above.Alternatively, maybe the problem expects an exact value in terms of exponents, but I think 10,159.36 is a reasonable approximate value.Wait, but let me see if I can express it as 100 * (2^10)^(2/3) = 100 * 2^(20/3) = 100 * 2^6 * 2^(2/3) = 100 * 64 * 2^(2/3) = 6400 * 2^(2/3). So, that's the exact form, but it's probably better to give a numerical approximation.Alternatively, maybe I can compute it more accurately.Compute 2^(2/3):We know that 2^(1/3) ≈ 1.25992105So, 2^(2/3) = (2^(1/3))^2 ≈ (1.25992105)^2 ≈ 1.58740105Therefore, 64 * 1.58740105 ≈ 64 * 1.58740105Compute 64 * 1.5 = 9664 * 0.08740105 ≈ 64 * 0.08 = 5.12; 64 * 0.00740105 ≈ 0.4736672So, total ≈ 5.12 + 0.4736672 ≈ 5.5936672Therefore, total ≈ 96 + 5.5936672 ≈ 101.5936672Therefore, a7 ≈ 100 * 101.5936672 ≈ 10,159.36672So, approximately 10,159.37 points.Alternatively, maybe I can write it as 10,159.37.But let me check if there's a better way to compute it.Alternatively, since a10 = 102,400 = 100 * r^9, so r^9 = 1024.Then, a7 = 100 * r^6 = 100 * (r^9)^(2/3) = 100 * (1024)^(2/3)Compute 1024^(2/3):1024 is 2^10, so 1024^(2/3) = (2^10)^(2/3) = 2^(20/3) = 2^6 * 2^(2/3) = 64 * 2^(2/3)As before, 2^(2/3) ≈ 1.5874, so 64 * 1.5874 ≈ 101.5936Therefore, a7 ≈ 100 * 101.5936 ≈ 10,159.36So, approximately 10,159.36 points.Alternatively, maybe the problem expects an exact value in terms of exponents, but I think the approximate value is acceptable.Wait, but let me see if 1024^(2/3) can be simplified.1024^(2/3) = (1024^(1/3))^2Compute 1024^(1/3):1024 is 10^3 is 1000, so 1024 is slightly more than 10^3. 10^3 = 1000, 10.079368^3 ≈ 1024.Wait, 10^3 = 100010.079368^3 ≈ 1024Wait, actually, 10.079368 is approximately the cube root of 1024.But 1024 is 2^10, so 1024^(1/3) = 2^(10/3) ≈ 2^3.3333 ≈ 10.079368Therefore, 1024^(2/3) = (1024^(1/3))^2 ≈ (10.079368)^2 ≈ 101.5936So, same result.Therefore, a7 ≈ 100 * 101.5936 ≈ 10,159.36So, approximately 10,159.36 points.Alternatively, maybe I can write it as 10,159.36, but perhaps the problem expects an exact value in terms of exponents, but I think the approximate decimal is fine.Alternatively, maybe the problem expects an exact value in terms of 2^(20/3), but that's probably not necessary.Alternatively, maybe I can express it as 100 * 2^(20/3), which is 100 * 2^6 * 2^(2/3) = 6400 * 2^(2/3). But again, that's not a whole number.Alternatively, maybe I can write it as 100 * (2^(10/9))^6 = 100 * 2^(60/9) = 100 * 2^(20/3), same as above.So, I think the approximate value is acceptable.Therefore, the common ratio is 2^(10/9), which is approximately 2.1602, and the 7th week's score is approximately 10,159.36 points.Wait, but let me check if 2^(10/9) is indeed the correct common ratio.Given that a10 = 100 * r^9 = 102,400So, r^9 = 1024Therefore, r = 1024^(1/9) = (2^10)^(1/9) = 2^(10/9). So, that's correct.Therefore, the common ratio is 2^(10/9), and the 7th term is 100 * (2^(10/9))^6 = 100 * 2^(60/9) = 100 * 2^(20/3) ≈ 10,159.36So, that seems correct.Alternatively, maybe I can write the 7th term as 100 * 2^(20/3), which is the exact form, but it's better to give a numerical approximation.Alternatively, maybe the problem expects an exact value in terms of exponents, but I think the approximate decimal is acceptable.Therefore, summarizing:1. The common difference of the arithmetic progression is 20/9 hours per week.2. The common ratio of the geometric progression is 2^(10/9), and the score in the 7th week is approximately 10,159.36 points.Wait, but let me check if 2^(10/9) is indeed the correct common ratio.Yes, because a10 = 100 * r^9 = 102,400, so r^9 = 1024, which is 2^10, so r = 2^(10/9). That's correct.And for the 7th term, a7 = 100 * r^6 = 100 * (2^(10/9))^6 = 100 * 2^(60/9) = 100 * 2^(20/3) ≈ 10,159.36Yes, that seems correct.So, I think that's the solution.</think>

🧠:<think>Okay, so I have this problem where a retiree is modeling noise levels with a sine function. The function is given as N(t) = A sin(Bt + C) + D. I need to find the constants A, B, C, D based on some given information, and then use that to find out how many hours a day the noise is above 65 dB.First, let me parse the information given. The maximum noise level is 70 dB at 8 AM, and the minimum is 50 dB at 8 PM. The periodicity is 24 hours. So, the function repeats every 24 hours.Starting with the first part: finding A, B, C, D.I remember that for a sine function of the form N(t) = A sin(Bt + C) + D, the amplitude is A, the period is 2π/B, the phase shift is -C/B, and D is the vertical shift, which is the average value.Given that the maximum is 70 and the minimum is 50, I can find A and D.The amplitude A is half the difference between the maximum and minimum. So, A = (70 - 50)/2 = 10. That makes sense because the sine function oscillates between -A and A, so adding D shifts it up.The vertical shift D is the average of the maximum and minimum. So, D = (70 + 50)/2 = 60. So, the function oscillates around 60 dB.Next, the period is 24 hours. The period of the sine function is 2π/B, so 2π/B = 24. Solving for B, we get B = 2π/24 = π/12. So, B is π/12.Now, we need to find C. The phase shift is given by -C/B. We know that the maximum occurs at 8 AM, which is t = 8 hours. Since the sine function reaches its maximum at π/2, we can set up the equation:Bt + C = π/2 when t = 8.So, substituting B = π/12, we have:(π/12)*8 + C = π/2Calculating (π/12)*8: that's (8π)/12 = (2π)/3.So, (2π)/3 + C = π/2Solving for C: C = π/2 - (2π)/3 = (3π/6 - 4π/6) = (-π)/6.So, C is -π/6.Let me double-check that. If I plug t = 8 into Bt + C, I should get π/2.(π/12)*8 + (-π/6) = (2π/3) - (π/6) = (4π/6 - π/6) = 3π/6 = π/2. Yes, that works.So, the function is N(t) = 10 sin((π/12)t - π/6) + 60.Wait, let me write that again: N(t) = 10 sin((π/12)t - π/6) + 60.Alternatively, it can be written as N(t) = 10 sin(π/12 (t - 2)) + 60, since factoring out π/12 from (π/12)t - π/6 gives π/12(t - 2). So, the phase shift is 2 hours, meaning the graph is shifted to the right by 2 hours. That makes sense because the maximum occurs at t = 8, which is 2 hours after t = 6, which would be the phase shift.Wait, actually, phase shift is -C/B, which was -(-π/6)/(π/12) = (π/6)/(π/12) = 2. So, the phase shift is 2 hours to the right. So, the function reaches maximum at t = 8, which is 2 hours after the usual maximum at t = 6. That seems correct.Okay, so I think I have A = 10, B = π/12, C = -π/6, D = 60.Now, moving on to part 2: finding the total duration when N(t) > 65 dB.So, we need to solve the inequality 10 sin((π/12)t - π/6) + 60 > 65.Subtracting 60 from both sides: 10 sin((π/12)t - π/6) > 5.Divide both sides by 10: sin((π/12)t - π/6) > 0.5.So, we need to find all t in [0, 24) where sin(θ) > 0.5, where θ = (π/12)t - π/6.I know that sin(θ) > 0.5 when θ is in (π/6 + 2πk, 5π/6 + 2πk) for integer k.So, let's solve for θ:π/6 < (π/12)t - π/6 < 5π/6Adding π/6 to all parts:π/6 + π/6 < (π/12)t < 5π/6 + π/6Simplify:π/3 < (π/12)t < πMultiply all parts by 12/π:(π/3)*(12/π) < t < π*(12/π)Simplify:4 < t < 12So, in the interval [0,24), the solution is t between 4 and 12, and also considering the periodicity, we might have another interval.Wait, but since the sine function is periodic with period 24, and we're looking within a 24-hour period, we need to check if there's another interval where sin(θ) > 0.5.But let's think: the sine function is above 0.5 in two intervals per period: one in the first half and one in the second half. Wait, but in our case, the period is 24 hours, so θ goes from -π/6 to (π/12)*24 - π/6 = 2π - π/6 = 11π/6.So, θ ranges from -π/6 to 11π/6.So, sin(θ) > 0.5 occurs when θ is in (π/6, 5π/6) and also when θ is in (13π/6, 17π/6), but since θ only goes up to 11π/6, the second interval would be from 13π/6 to 11π/6, but 13π/6 is greater than 11π/6, so actually, there's only one interval in θ where sin(θ) > 0.5, which is (π/6, 5π/6).Wait, that can't be right because in a full period, sine is above 0.5 in two intervals. Maybe I need to consider the entire θ range.Wait, θ starts at -π/6 when t=0, and goes up to 11π/6 when t=24.So, in the θ interval from -π/6 to 11π/6, sin(θ) > 0.5 occurs in two places: once in the first half and once in the second half.Wait, let me plot θ from -π/6 to 11π/6. So, starting at -π/6, going up to 11π/6, which is equivalent to going from -30 degrees to 330 degrees.In this range, sin(θ) > 0.5 occurs in two intervals:1. From θ = π/6 (30 degrees) to θ = 5π/6 (150 degrees).2. From θ = 13π/6 (390 degrees, which is equivalent to 30 degrees) to θ = 17π/6 (340 degrees, which is equivalent to 170 degrees). But wait, 13π/6 is beyond 2π, which is 12π/6. So, 13π/6 is π/6 beyond 2π, which is the same as π/6. Similarly, 17π/6 is 5π/6 beyond 2π.But in our θ range, we only go up to 11π/6, which is 330 degrees. So, the second interval where sin(θ) > 0.5 would be from θ = 13π/6 - 2π = π/6 to θ = 17π/6 - 2π = 5π/6. Wait, that doesn't make sense because 13π/6 is beyond 2π.Wait, perhaps I should think of θ in terms of modulo 2π. So, θ ranges from -π/6 to 11π/6, which is the same as from 11π/6 - 2π = -π/6 to 11π/6.So, in terms of 0 to 2π, θ from 0 to 11π/6 is equivalent to θ from 0 to 2π minus π/6.So, sin(θ) > 0.5 occurs in two intervals:1. From π/6 to 5π/6.2. From 13π/6 to 17π/6, but since 13π/6 is equivalent to π/6 (since 13π/6 - 2π = π/6), and 17π/6 is equivalent to 5π/6 (17π/6 - 2π = 5π/6). So, in the range from 0 to 11π/6, the second interval would be from 13π/6 - 2π = π/6 to 17π/6 - 2π = 5π/6, but since 13π/6 is beyond 2π, which is 12π/6, so 13π/6 is π/6 beyond 2π. But our θ only goes up to 11π/6, which is 330 degrees, so the second interval where sin(θ) > 0.5 would be from θ = 13π/6 - 2π = π/6 to θ = 17π/6 - 2π = 5π/6, but since 13π/6 is beyond our θ range, we need to see if θ reaches 13π/6.Wait, θ at t=24 is 11π/6, which is less than 13π/6. So, in our θ range, sin(θ) > 0.5 only occurs once, from π/6 to 5π/6.Wait, that can't be right because in a full period, sine is above 0.5 twice. Maybe I'm making a mistake here.Alternatively, perhaps I should solve the inequality without worrying about the θ range, but instead solve for t directly.So, sin((π/12)t - π/6) > 0.5.The general solution for sin(x) > 0.5 is x ∈ (π/6 + 2πk, 5π/6 + 2πk) for integer k.So, let's set x = (π/12)t - π/6.So, (π/12)t - π/6 ∈ (π/6 + 2πk, 5π/6 + 2πk).Solving for t:Add π/6 to all parts:(π/12)t ∈ (π/6 + π/6 + 2πk, 5π/6 + π/6 + 2πk)Simplify:(π/12)t ∈ (π/3 + 2πk, π + 2πk)Multiply all parts by 12/π:t ∈ ( (π/3)*(12/π) + (2πk)*(12/π), (π)*(12/π) + (2πk)*(12/π) )Simplify:t ∈ (4 + 24k, 12 + 24k)Since we're looking for t in [0,24), k can be 0 or 1.For k=0: t ∈ (4,12)For k=1: t ∈ (28,36), but since 28 >24, we ignore this.So, the solution is t ∈ (4,12). So, the noise level is above 65 dB from 4 AM to 12 PM, which is 8 hours.Wait, but that seems too long. Let me check.Wait, if the maximum is at 8 AM, which is t=8, and the function is symmetric, then the noise level should be above 65 dB for some time before and after the maximum.But according to the solution, it's above 65 dB from 4 AM to 12 PM, which is 8 hours. Let me verify this.Let me plug in t=4: N(4) = 10 sin((π/12)*4 - π/6) + 60.Calculate the argument: (π/12)*4 = π/3, so π/3 - π/6 = π/6. sin(π/6) = 0.5, so N(4) = 10*0.5 + 60 = 5 + 60 = 65 dB.Similarly, at t=12: (π/12)*12 = π, so π - π/6 = 5π/6. sin(5π/6) = 0.5, so N(12) = 10*0.5 + 60 = 65 dB.So, at t=4 and t=12, the noise is exactly 65 dB. So, the noise is above 65 dB between t=4 and t=12, which is 8 hours.Wait, but intuitively, if the maximum is at t=8, and the function is symmetric, the time above 65 dB should be symmetric around t=8. So, from t=4 to t=12 is 8 hours, which is symmetric around t=8.So, that seems correct.Therefore, the total duration is 8 hours.Wait, but let me think again. The sine function is above 0.5 for two intervals in each period, but in this case, since the period is 24 hours, but the function is shifted, maybe it only crosses 0.5 once? No, no, the general solution for sin(x) > 0.5 is two intervals per period, but in our case, because of the phase shift, it's only crossing once in the 24-hour period? No, that can't be.Wait, no, the general solution is that in each period, sin(x) is above 0.5 for two intervals. But in our case, because of the phase shift, the interval where sin(x) > 0.5 is shifted, but it's still two intervals.Wait, but in our solution, we only found one interval, from t=4 to t=12. That's 8 hours. But shouldn't it be two intervals, each of 4 hours, totaling 8 hours?Wait, no, because in the general solution, each period has two intervals where sin(x) > 0.5, each of length (5π/6 - π/6) = 4π/6 = 2π/3, which is 120 degrees, or 1/3 of the period. So, in a 24-hour period, each interval is 8 hours (since 24*(2π/3)/(2π)) = 24*(1/3) = 8 hours. Wait, that doesn't make sense.Wait, no, the length of each interval where sin(x) > 0.5 is (5π/6 - π/6) = 4π/6 = 2π/3 radians. Since the period is 2π, the fraction is (2π/3)/(2π) = 1/3. So, in a 24-hour period, the total time where sin(x) > 0.5 is 24*(1/3) = 8 hours, which is consistent with our solution.So, the noise level is above 65 dB for 8 hours a day.Wait, but let me double-check by plugging in t=0: N(0) = 10 sin(-π/6) + 60 = 10*(-0.5) + 60 = -5 + 60 = 55 dB, which is below 65.At t=8: N(8) = 10 sin((π/12)*8 - π/6) + 60 = 10 sin(2π/3 - π/6) + 60 = 10 sin(π/2) + 60 = 10*1 + 60 = 70 dB.At t=4: N(4) = 65 dB, as calculated before.At t=12: N(12) = 65 dB.So, from t=4 to t=12, the noise is above 65 dB, which is 8 hours.Therefore, the total duration is 8 hours.I think that's correct.</think>

🧠:<think>Alright, so I have this problem about figure skating, which is pretty cool. It's divided into two parts: trajectory analysis and angular momentum. Let me tackle them one by one.Starting with the first part: Trajectory Analysis. The problem gives me parametric equations for the jump:x(t) = 3t,y(t) = 4t - 5t²,z(t) = 2t².I need to find the total airtime and the maximum height reached. Hmm, okay. So, airtime would be the duration the skater is in the air, right? That would be from when they leave the ground until they land again. Since these are parametric equations, I guess I need to figure out when the skater is on the ground, which is when z(t) = 0 because that's the height.Looking at z(t) = 2t². When does z(t) = 0? Well, 2t² = 0 implies t = 0. Hmm, that's just the start. But wait, maybe the skater lands when z(t) = 0 again after takeoff. But z(t) is 2t², which is a parabola opening upwards, so it only touches zero at t=0. That doesn't make sense because the skater should leave the ground, go up, and come back down. Maybe I'm missing something here.Wait, perhaps the equations are only modeling the jump after takeoff, so t=0 is when the skater is already in the air. So, to find when they land, I need to see if z(t) ever returns to zero. But z(t) = 2t² is always non-negative and only zero at t=0. That doesn't seem right. Maybe I made a mistake.Wait, let me check the equations again. x(t) = 3t, y(t) = 4t - 5t², z(t) = 2t². Hmm, z(t) is 2t², which is zero only at t=0. So, does that mean the skater never lands? That can't be. Maybe the equations are only for the ascent, and the descent is modeled differently? Or perhaps there's a typo in the problem?Wait, maybe I need to consider that the skater starts on the ground, jumps, reaches a maximum height, and then comes back down. So, perhaps z(t) should be a quadratic that goes up and then comes back down. But z(t) = 2t² is a parabola opening upwards, so it just keeps increasing. That doesn't make sense for a jump. Maybe it's supposed to be z(t) = -2t² + something? Or maybe it's z(t) = 2t² - something?Wait, let me think. If z(t) is the height, then it should have a maximum and then come back down. So, maybe the equation is supposed to be z(t) = -2t² + something. But in the problem, it's given as z(t) = 2t². Hmm, that's confusing.Wait, maybe the skater is jumping on the moon or something where gravity is different? But no, the equations are given, so I have to work with them. Maybe the skater doesn't land? That seems odd.Wait, perhaps the equations are given for the entire trajectory, but z(t) is 2t², which is always increasing. So, maybe the skater never lands? That can't be. Maybe the equations are only for the ascent? But then, how do we find the total airtime?Wait, maybe I need to consider that the skater starts at t=0, jumps, and then comes back down at some t where z(t) = 0 again. But since z(t) = 2t², the only solution is t=0. So, perhaps the equations are only modeling the ascent, and the descent is symmetric? But without more information, I can't assume that.Wait, maybe I'm overcomplicating this. Let me check the equations again. x(t) = 3t, y(t) = 4t - 5t², z(t) = 2t². So, z(t) is 2t², which is a parabola opening upwards. So, the skater starts at (0,0,0), and as time increases, z increases. So, does that mean the skater is continuously ascending? That doesn't make sense for a jump.Wait, maybe the equations are given in a different coordinate system. Maybe z(t) is not the vertical component but something else? Or perhaps the skater is on a moving platform? Hmm, I'm not sure.Wait, maybe the problem is just modeling the jump in a way that z(t) is the height, but it's given as 2t², which is always increasing. So, perhaps the skater never lands, which is impossible. Maybe the equations are incorrect? Or maybe I'm misinterpreting them.Wait, let me think differently. Maybe the skater is in the air from t=0 to t=T, and z(T) = 0 again. But z(t) = 2t², so 2T² = 0 implies T=0, which is the same as the start. So, that can't be.Wait, maybe the equations are given in a way that z(t) is the height above the takeoff point, so when the skater lands, z(t) = 0 again. But z(t) = 2t², so that would only happen at t=0. So, maybe the equations are only for the ascent, and the descent is not modeled here.But then, how do I find the total airtime? Maybe I need to find when the skater lands, but since z(t) never returns to zero, perhaps the equations are incomplete or incorrect.Wait, maybe I made a mistake in interpreting the equations. Let me check again. x(t) = 3t, y(t) = 4t - 5t², z(t) = 2t². So, x is linear, y is quadratic, z is quadratic. Maybe z(t) is supposed to be the height, but it's given as 2t², which is always increasing. That's odd.Wait, maybe the skater is jumping on a trampoline or something where they go up and then come back down, but the equations don't model that. Hmm.Wait, perhaps the equations are given for the entire trajectory, but z(t) is 2t², which is always increasing. So, maybe the skater is in the air indefinitely? That can't be.Wait, maybe I need to consider that the skater starts at t=0, jumps, and the equations are given for the time in the air, but z(t) is 2t², so the skater is always ascending. That doesn't make sense.Wait, maybe the equations are given in a different way. Maybe z(t) is the height, but it's given as 2t², so the skater is ascending with increasing height. So, perhaps the skater never lands, which is impossible. Maybe the problem is miswritten.Wait, maybe I need to consider that the skater starts at t=0, jumps, and the equations are given for the entire trajectory, but z(t) is 2t², which is always increasing. So, maybe the skater is in the air forever? That can't be.Wait, maybe I'm overcomplicating this. Let me think about the problem again. It says "represent the trajectory of a jump as a parametric equation." So, maybe the equations are correct, and I just need to find the total airtime and maximum height.But if z(t) = 2t², then the skater is always ascending, which is not possible. So, maybe the equations are given incorrectly. Maybe it's supposed to be z(t) = -2t² + something? Or maybe z(t) = 2t - 5t²? That would make more sense because it would have a maximum and then come back down.Wait, let me check the problem again. It says z(t) = 2t². Hmm. Maybe I need to proceed with that, even though it seems odd.So, if z(t) = 2t², then the skater is always ascending. So, the maximum height would be at the end of the jump, but since it never lands, the maximum height is at the end of the airtime. But since the airtime is infinite, that doesn't make sense.Wait, maybe the problem is given in a way that the skater is in the air for a certain time, and z(t) is 2t², so the maximum height is at the end of the airtime. But without knowing when the skater lands, I can't find the airtime.Wait, maybe the skater lands when y(t) returns to zero? Because y(t) = 4t - 5t². So, when does y(t) = 0?Solving 4t - 5t² = 0.t(4 - 5t) = 0.So, t=0 or t=4/5 seconds.So, maybe the skater lands at t=4/5 seconds. That would make sense because y(t) is the horizontal displacement, and when y(t) returns to zero, the skater has completed the jump.So, total airtime would be t=4/5 seconds.Okay, that seems plausible. So, even though z(t) is always increasing, maybe the skater lands when y(t) returns to zero, which is at t=4/5.So, total airtime is 4/5 seconds.Now, for the maximum height, since z(t) = 2t², and the skater is in the air until t=4/5, the maximum height would be at t=4/5.So, z(4/5) = 2*(4/5)^2 = 2*(16/25) = 32/25 = 1.28 meters.Wait, but z(t) is increasing throughout, so the maximum height is indeed at t=4/5, which is 1.28 meters.Okay, that seems to make sense. So, even though z(t) is always increasing, the skater lands when y(t) returns to zero, which is at t=4/5. So, the total airtime is 4/5 seconds, and the maximum height is 1.28 meters.Alright, moving on to the second part: Angular Momentum.The problem says that a skater pulls in their arms to increase rotational speed. Initially, the moment of inertia is I₁ = 2.5 kg·m², and angular velocity ω₁ = 5 rad/s. After pulling in arms, I₂ = 1.0 kg·m². We need to find ω₂ using conservation of angular momentum, and then find the kinetic energy before and after.Okay, so angular momentum L is conserved, so L₁ = L₂.Angular momentum L = Iω.So, I₁ω₁ = I₂ω₂.So, ω₂ = (I₁/I₂) * ω₁.Plugging in the numbers:ω₂ = (2.5 / 1.0) * 5 = 2.5 * 5 = 12.5 rad/s.Okay, so the new angular velocity is 12.5 rad/s.Now, kinetic energy before and after.Kinetic energy for rotation is (1/2)Iω².So, initial kinetic energy KE₁ = (1/2) * I₁ * ω₁².Plugging in:KE₁ = 0.5 * 2.5 * (5)^2 = 0.5 * 2.5 * 25 = 0.5 * 62.5 = 31.25 J.After pulling in arms, KE₂ = (1/2) * I₂ * ω₂².Plugging in:KE₂ = 0.5 * 1.0 * (12.5)^2 = 0.5 * 1.0 * 156.25 = 0.5 * 156.25 = 78.125 J.So, the kinetic energy increases from 31.25 J to 78.125 J when the skater pulls in their arms.Wait, that makes sense because when the skater pulls in their arms, they reduce their moment of inertia, which increases their angular velocity, and since kinetic energy depends on the square of angular velocity, the kinetic energy increases.Alright, so summarizing:1. Total airtime is 4/5 seconds, maximum height is 1.28 meters.2. New angular velocity is 12.5 rad/s, initial KE is 31.25 J, final KE is 78.125 J.I think that's it. Let me just double-check my calculations.For the trajectory:- Airtime: solved y(t)=0, got t=0 and t=4/5. So, airtime is 4/5 seconds.- Maximum height: z(4/5) = 2*(16/25) = 32/25 = 1.28 meters.For angular momentum:- Conservation: I₁ω₁ = I₂ω₂ => ω₂ = (2.5/1.0)*5 = 12.5 rad/s.- KE₁ = 0.5*2.5*25 = 31.25 J.- KE₂ = 0.5*1.0*156.25 = 78.125 J.Yes, that all checks out.</think>

🧠:<think>Okay, so I need to figure out an explicit formula for the sequence ( a_n ) defined by the recursive relationship:[a_{n+1} = frac{1}{2}a_n + frac{1}{2}(b_n + c_n)]where ( b_n = 3 sin(n) ) and ( c_n = 2 cos(n) ). The initial condition is ( a_0 = 10 ). Then, I also need to determine if this sequence converges as ( n ) approaches infinity and find its limit if it does.First, let me try to understand the recursion. It seems that each term ( a_{n+1} ) is the average of the previous term ( a_n ) and the sum of ( b_n ) and ( c_n ). So, it's a linear recurrence relation with some trigonometric functions involved.I remember that linear recursions can sometimes be solved by finding the homogeneous solution and a particular solution. The general form of such a recurrence is:[a_{n+1} = k a_n + f(n)]In this case, ( k = frac{1}{2} ) and ( f(n) = frac{1}{2}(b_n + c_n) = frac{1}{2}(3 sin(n) + 2 cos(n)) ).So, the homogeneous equation is:[a_{n+1} = frac{1}{2} a_n]The solution to this is straightforward. It's a geometric sequence where each term is half of the previous one. So, the homogeneous solution ( a_n^{(h)} ) is:[a_n^{(h)} = C left( frac{1}{2} right)^n]where ( C ) is a constant determined by initial conditions.Now, I need to find a particular solution ( a_n^{(p)} ) to the nonhomogeneous equation. Since the nonhomogeneous term ( f(n) ) is a combination of sine and cosine functions, I can try a particular solution of the form:[a_n^{(p)} = A sin(n) + B cos(n)]where ( A ) and ( B ) are constants to be determined.Let me plug ( a_n^{(p)} ) into the recurrence relation:[a_{n+1}^{(p)} = frac{1}{2} a_n^{(p)} + frac{1}{2}(3 sin(n) + 2 cos(n))]First, compute ( a_{n+1}^{(p)} ):[a_{n+1}^{(p)} = A sin(n+1) + B cos(n+1)]Using the trigonometric identities:[sin(n+1) = sin(n)cos(1) + cos(n)sin(1)][cos(n+1) = cos(n)cos(1) - sin(n)sin(1)]So,[a_{n+1}^{(p)} = A [sin(n)cos(1) + cos(n)sin(1)] + B [cos(n)cos(1) - sin(n)sin(1)]][= A cos(1) sin(n) + A sin(1) cos(n) + B cos(1) cos(n) - B sin(1) sin(n)][= [A cos(1) - B sin(1)] sin(n) + [A sin(1) + B cos(1)] cos(n)]Now, the right-hand side of the recurrence is:[frac{1}{2} a_n^{(p)} + frac{1}{2}(3 sin(n) + 2 cos(n))][= frac{1}{2} [A sin(n) + B cos(n)] + frac{3}{2} sin(n) + cos(n)][= left( frac{A}{2} + frac{3}{2} right) sin(n) + left( frac{B}{2} + 1 right) cos(n)]So, equating the coefficients of ( sin(n) ) and ( cos(n) ) from both sides:For ( sin(n) ):[A cos(1) - B sin(1) = frac{A}{2} + frac{3}{2}]For ( cos(n) ):[A sin(1) + B cos(1) = frac{B}{2} + 1]So, we have a system of two equations:1. ( A cos(1) - B sin(1) = frac{A}{2} + frac{3}{2} )2. ( A sin(1) + B cos(1) = frac{B}{2} + 1 )Let me rewrite these equations:1. ( A cos(1) - B sin(1) - frac{A}{2} - frac{3}{2} = 0 )2. ( A sin(1) + B cos(1) - frac{B}{2} - 1 = 0 )Simplify equation 1:Factor out A and B:( A ( cos(1) - frac{1}{2} ) - B sin(1) - frac{3}{2} = 0 )Similarly, equation 2:( A sin(1) + B ( cos(1) - frac{1}{2} ) - 1 = 0 )So, now we have:1. ( A ( cos(1) - frac{1}{2} ) - B sin(1) = frac{3}{2} )2. ( A sin(1) + B ( cos(1) - frac{1}{2} ) = 1 )Let me denote ( C = cos(1) - frac{1}{2} ) and ( D = sin(1) ) to simplify the notation.Then, the system becomes:1. ( A C - B D = frac{3}{2} )2. ( A D + B C = 1 )This is a linear system in variables A and B. Let me write it in matrix form:[begin{pmatrix}C & -D D & Cend{pmatrix}begin{pmatrix}A Bend{pmatrix}=begin{pmatrix}frac{3}{2} 1end{pmatrix}]To solve this, I can compute the determinant of the coefficient matrix:Determinant ( Delta = C cdot C - (-D) cdot D = C^2 + D^2 )Compute ( C^2 + D^2 ):Since ( C = cos(1) - frac{1}{2} ) and ( D = sin(1) ),( C^2 + D^2 = (cos(1) - frac{1}{2})^2 + sin^2(1) )( = cos^2(1) - cos(1) + frac{1}{4} + sin^2(1) )( = (cos^2(1) + sin^2(1)) - cos(1) + frac{1}{4} )( = 1 - cos(1) + frac{1}{4} )( = frac{5}{4} - cos(1) )So, determinant ( Delta = frac{5}{4} - cos(1) ). Since ( cos(1) ) is approximately 0.5403, so ( Delta approx frac{5}{4} - 0.5403 approx 1.25 - 0.5403 = 0.7097 ), which is positive, so the system has a unique solution.Using Cramer's rule, the solution is:( A = frac{ Delta_A }{ Delta } )( B = frac{ Delta_B }{ Delta } )Where ( Delta_A ) is the determinant when replacing the first column with the constants, and ( Delta_B ) is the determinant when replacing the second column.Compute ( Delta_A ):Replace first column with ( frac{3}{2} ) and 1:[Delta_A = begin{vmatrix}frac{3}{2} & -D 1 & Cend{vmatrix}= frac{3}{2} C - (-D)(1) = frac{3}{2} C + D]Similarly, ( Delta_B ):Replace second column with ( frac{3}{2} ) and 1:[Delta_B = begin{vmatrix}C & frac{3}{2} D & 1end{vmatrix}= C cdot 1 - frac{3}{2} D = C - frac{3}{2} D]So,( A = frac{ frac{3}{2} C + D }{ Delta } )( B = frac{ C - frac{3}{2} D }{ Delta } )Now, substitute back ( C = cos(1) - frac{1}{2} ) and ( D = sin(1) ):Compute ( frac{3}{2} C + D ):( frac{3}{2} (cos(1) - frac{1}{2}) + sin(1) )( = frac{3}{2} cos(1) - frac{3}{4} + sin(1) )Compute ( C - frac{3}{2} D ):( (cos(1) - frac{1}{2}) - frac{3}{2} sin(1) )( = cos(1) - frac{1}{2} - frac{3}{2} sin(1) )So,( A = frac{ frac{3}{2} cos(1) - frac{3}{4} + sin(1) }{ frac{5}{4} - cos(1) } )( B = frac{ cos(1) - frac{1}{2} - frac{3}{2} sin(1) }{ frac{5}{4} - cos(1) } )This looks complicated, but maybe we can factor out some terms or simplify.Let me compute the numerator and denominator for A:Numerator A:( frac{3}{2} cos(1) + sin(1) - frac{3}{4} )Denominator:( frac{5}{4} - cos(1) )Similarly for B:Numerator B:( cos(1) - frac{1}{2} - frac{3}{2} sin(1) )Denominator same as above.Hmm, perhaps we can factor out 1/4 in the denominator:Denominator: ( frac{5}{4} - cos(1) = frac{5 - 4 cos(1)}{4} )Similarly, let's express numerator A over 4:Numerator A:( frac{3}{2} cos(1) + sin(1) - frac{3}{4} = frac{6 cos(1) + 4 sin(1) - 3}{4} )Similarly, numerator B:( cos(1) - frac{1}{2} - frac{3}{2} sin(1) = frac{2 cos(1) - 1 - 6 sin(1)}{4} )Wait, let me check:For numerator A:( frac{3}{2} cos(1) = frac{6}{4} cos(1) )( sin(1) = frac{4}{4} sin(1) )( - frac{3}{4} = - frac{3}{4} )So, altogether:( frac{6 cos(1) + 4 sin(1) - 3}{4} )Similarly, numerator B:( cos(1) = frac{4}{4} cos(1) )( - frac{1}{2} = - frac{2}{4} )( - frac{3}{2} sin(1) = - frac{6}{4} sin(1) )So,( frac{4 cos(1) - 2 - 6 sin(1)}{4} )Therefore, now:( A = frac{ frac{6 cos(1) + 4 sin(1) - 3}{4} }{ frac{5 - 4 cos(1)}{4} } = frac{6 cos(1) + 4 sin(1) - 3}{5 - 4 cos(1)} )Similarly,( B = frac{ frac{4 cos(1) - 2 - 6 sin(1)}{4} }{ frac{5 - 4 cos(1)}{4} } = frac{4 cos(1) - 2 - 6 sin(1)}{5 - 4 cos(1)} )So, now we have expressions for A and B:( A = frac{6 cos(1) + 4 sin(1) - 3}{5 - 4 cos(1)} )( B = frac{4 cos(1) - 2 - 6 sin(1)}{5 - 4 cos(1)} )These expressions can be simplified further, perhaps by factoring numerator and denominator.Looking at the denominator: ( 5 - 4 cos(1) ). The numerator for A:( 6 cos(1) + 4 sin(1) - 3 )I wonder if this can be expressed as a multiple of the denominator or something similar, but it's not obvious.Alternatively, perhaps we can factor numerator and denominator:Let me compute the numerator for A:( 6 cos(1) + 4 sin(1) - 3 )Let me see if this can be written as ( k (5 - 4 cos(1)) + m ), but that might not help.Alternatively, perhaps we can factor out a 2:( 6 cos(1) - 3 = 3(2 cos(1) - 1) )So,( 6 cos(1) + 4 sin(1) - 3 = 3(2 cos(1) - 1) + 4 sin(1) )Similarly, the denominator is ( 5 - 4 cos(1) ). Hmm, perhaps not helpful.Alternatively, perhaps we can write the numerator as a linear combination of the denominator and something else.Wait, maybe it's better to just leave it as is for now.So, the particular solution is:[a_n^{(p)} = A sin(n) + B cos(n)]with A and B as above.Therefore, the general solution is the sum of the homogeneous and particular solutions:[a_n = a_n^{(h)} + a_n^{(p)} = C left( frac{1}{2} right)^n + A sin(n) + B cos(n)]Now, we need to find the constant C using the initial condition ( a_0 = 10 ).Compute ( a_0 ):[a_0 = C left( frac{1}{2} right)^0 + A sin(0) + B cos(0)][= C cdot 1 + A cdot 0 + B cdot 1][= C + B = 10]So, ( C = 10 - B )Therefore, the explicit formula is:[a_n = (10 - B) left( frac{1}{2} right)^n + A sin(n) + B cos(n)]Now, substitute A and B:But since A and B are expressed in terms of trigonometric functions, this might get complicated. Alternatively, perhaps we can write the entire expression in terms of sine and cosine with some amplitude and phase shift, but that might not be necessary unless asked.Alternatively, perhaps we can express the solution as:[a_n = (10 - B) left( frac{1}{2} right)^n + A sin(n) + B cos(n)]But since A and B are constants, this is an explicit formula.However, perhaps we can write this in a more compact form. Let me see.Alternatively, since the homogeneous solution decays to zero as ( n ) increases because ( left( frac{1}{2} right)^n ) tends to zero, the long-term behavior is dominated by the particular solution ( A sin(n) + B cos(n) ). So, the sequence ( a_n ) will oscillate with decreasing amplitude from the homogeneous part and a steady oscillation from the particular solution.But let's get back to the explicit formula.Given that ( C = 10 - B ), and B is:( B = frac{4 cos(1) - 2 - 6 sin(1)}{5 - 4 cos(1)} )So,( C = 10 - frac{4 cos(1) - 2 - 6 sin(1)}{5 - 4 cos(1)} )Let me compute this:( C = frac{10 (5 - 4 cos(1)) - (4 cos(1) - 2 - 6 sin(1))}{5 - 4 cos(1)} )Compute numerator:( 10(5 - 4 cos(1)) = 50 - 40 cos(1) )Subtract ( (4 cos(1) - 2 - 6 sin(1)) ):( 50 - 40 cos(1) - 4 cos(1) + 2 + 6 sin(1) )( = 52 - 44 cos(1) + 6 sin(1) )So,( C = frac{52 - 44 cos(1) + 6 sin(1)}{5 - 4 cos(1)} )Therefore, the explicit formula is:[a_n = left( frac{52 - 44 cos(1) + 6 sin(1)}{5 - 4 cos(1)} right) left( frac{1}{2} right)^n + left( frac{6 cos(1) + 4 sin(1) - 3}{5 - 4 cos(1)} right) sin(n) + left( frac{4 cos(1) - 2 - 6 sin(1)}{5 - 4 cos(1)} right) cos(n)]This is quite a complex expression, but it is explicit.Alternatively, perhaps we can factor out ( frac{1}{5 - 4 cos(1)} ) from all terms:[a_n = frac{1}{5 - 4 cos(1)} left[ (52 - 44 cos(1) + 6 sin(1)) left( frac{1}{2} right)^n + (6 cos(1) + 4 sin(1) - 3) sin(n) + (4 cos(1) - 2 - 6 sin(1)) cos(n) right]]This might be a cleaner way to present it.Now, moving on to part 2: determining if ( a_n ) converges as ( n ) approaches infinity.Looking at the explicit formula, we have three terms:1. ( (10 - B) left( frac{1}{2} right)^n ): This term decays exponentially to zero as ( n ) increases because ( frac{1}{2} < 1 ).2. ( A sin(n) ): This term oscillates between ( -|A| ) and ( |A| ).3. ( B cos(n) ): Similarly, this term oscillates between ( -|B| ) and ( |B| ).Therefore, as ( n ) approaches infinity, the first term goes to zero, and the remaining terms oscillate. So, the sequence ( a_n ) does not converge to a single value because it continues to oscillate. However, the amplitude of the oscillation is determined by the coefficients A and B.But wait, let me think again. The homogeneous solution decays, so the influence of the initial condition diminishes over time, and the particular solution remains. So, does the sequence converge? Or does it oscillate indefinitely?In the explicit formula, as ( n to infty ), ( a_n ) approaches ( A sin(n) + B cos(n) ), which is a bounded oscillating function. Therefore, the sequence does not converge to a single limit; instead, it oscillates within a fixed range.But wait, let me check if ( A ) and ( B ) are such that ( A sin(n) + B cos(n) ) can be expressed as a single sinusoidal function with some amplitude. The amplitude would be ( sqrt{A^2 + B^2} ), so the sequence oscillates between ( -sqrt{A^2 + B^2} ) and ( sqrt{A^2 + B^2} ).But since ( A ) and ( B ) are constants, the oscillations are persistent and do not die out. Therefore, the sequence does not converge; it oscillates indefinitely.However, let me verify this conclusion because sometimes even with oscillatory terms, if their coefficients decay, the sequence might converge. But in this case, the particular solution is a steady oscillation without decay, so the entire sequence will oscillate without settling down.Therefore, the sequence does not converge as ( n ) approaches infinity.But wait, let me think again. The homogeneous solution is decaying, so the influence of the initial condition diminishes, but the particular solution is a steady oscillation. So, the overall behavior is that ( a_n ) approaches the particular solution as ( n ) increases, but since the particular solution is oscillatory, the sequence does not converge to a single value.Hence, the sequence does not converge; it oscillates with a fixed amplitude determined by ( A ) and ( B ).But to be thorough, let me compute the limit as ( n to infty ):[lim_{n to infty} a_n = lim_{n to infty} left[ (10 - B) left( frac{1}{2} right)^n + A sin(n) + B cos(n) right]]The first term goes to zero, and the remaining terms oscillate. Since sine and cosine are bounded but do not approach any limit, the overall limit does not exist. Therefore, the sequence does not converge.Alternatively, if we consider the average behavior or something else, but in terms of the standard limit, it does not exist.So, to summarize:1. The explicit formula for ( a_n ) is:[a_n = frac{1}{5 - 4 cos(1)} left[ (52 - 44 cos(1) + 6 sin(1)) left( frac{1}{2} right)^n + (6 cos(1) + 4 sin(1) - 3) sin(n) + (4 cos(1) - 2 - 6 sin(1)) cos(n) right]]2. The sequence ( a_n ) does not converge as ( n ) approaches infinity because it oscillates indefinitely.</think>