💡Alright, so I have this problem about an arithmetic sequence. Let me try to understand it step by step. The problem says:"Given an arithmetic sequence {a_n}, where S_3 = 42, S_6 = 57, determine a_n = ______, and when S_n takes the maximum value, n = ______."Okay, so I need to find the general term of the arithmetic sequence, a_n, and also find the value of n where the sum S_n is maximized.First, let me recall what an arithmetic sequence is. An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference, usually denoted by d. The general term of an arithmetic sequence can be written as:a_n = a_1 + (n - 1)dwhere a_1 is the first term.Now, the sum of the first n terms of an arithmetic sequence is given by the formula:S_n = n/2 * [2a_1 + (n - 1)d]Alternatively, this can also be written as:S_n = n * (a_1 + a_n)/2Since we're given S_3 and S_6, I can set up two equations using the sum formula and solve for a_1 and d.Let me write down the given information:S_3 = 42S_6 = 57So, plugging n = 3 into the sum formula:S_3 = 3/2 * [2a_1 + (3 - 1)d] = 42Simplify that:3/2 * [2a_1 + 2d] = 42Multiply both sides by 2 to eliminate the fraction:3 * [2a_1 + 2d] = 84Divide both sides by 3:2a_1 + 2d = 28I can simplify this further by dividing both sides by 2:a_1 + d = 14 ...(1)Now, let's do the same for S_6:S_6 = 6/2 * [2a_1 + (6 - 1)d] = 57Simplify:3 * [2a_1 + 5d] = 57Divide both sides by 3:2a_1 + 5d = 19 ...(2)Now, I have two equations:1) a_1 + d = 142) 2a_1 + 5d = 19I can solve these simultaneously. Let me solve equation (1) for a_1:a_1 = 14 - dNow, substitute this into equation (2):2*(14 - d) + 5d = 19Multiply out:28 - 2d + 5d = 19Combine like terms:28 + 3d = 19Subtract 28 from both sides:3d = -9Divide both sides by 3:d = -3Now that I have d, I can find a_1 using equation (1):a_1 + (-3) = 14So,a_1 = 14 + 3 = 17Alright, so the first term a_1 is 17, and the common difference d is -3.Now, let's write the general term a_n:a_n = a_1 + (n - 1)d= 17 + (n - 1)*(-3)= 17 - 3(n - 1)= 17 - 3n + 3= 20 - 3nSo, a_n = 20 - 3n.Now, the second part of the problem asks for the value of n where S_n is maximized.To find the maximum sum, I need to analyze the behavior of S_n. Since this is an arithmetic sequence with a negative common difference (d = -3), the terms are decreasing. That means the sequence starts at 17 and each subsequent term is 3 less than the previous one.Eventually, the terms will become negative. The sum S_n will increase as long as the terms are positive, but once the terms become negative, adding them will start decreasing the total sum. Therefore, the maximum sum occurs just before the terms become negative.So, I need to find the largest n for which a_n is still non-negative.Set a_n >= 0:20 - 3n >= 0Solve for n:20 >= 3nn <= 20/3n <= 6.666...Since n must be an integer (as it's the term number), the largest integer less than or equal to 6.666 is 6.Therefore, the sum S_n reaches its maximum value when n = 6.Let me double-check this by calculating S_6 and S_7 to ensure that S_6 is indeed larger than S_7.First, calculate S_6:We already know S_6 = 57.Now, calculate S_7:S_7 = S_6 + a_7We know a_7 = 20 - 3*7 = 20 - 21 = -1So, S_7 = 57 + (-1) = 56Indeed, S_7 is less than S_6, confirming that the maximum sum occurs at n = 6.Just to be thorough, let me calculate a few terms and the sums to see the trend.Compute a_1 to a_7:a_1 = 17a_2 = 14a_3 = 11a_4 = 8a_5 = 5a_6 = 2a_7 = -1Now, compute S_1 to S_7:S_1 = 17S_2 = 17 + 14 = 31S_3 = 31 + 11 = 42S_4 = 42 + 8 = 50S_5 = 50 + 5 = 55S_6 = 55 + 2 = 57S_7 = 57 + (-1) = 56Looking at the sums: 17, 31, 42, 50, 55, 57, 56.Yes, the sum increases up to n=6 and then decreases at n=7. So, n=6 is indeed where the maximum sum occurs.Therefore, the answers are:a_n = 20 - 3nandn = 6**Final Answer**The general term is boxed{20 - 3n} and the value of ( n ) where ( S_n ) is maximized is boxed{6}.

💡First, I need to determine the 4500th positive even integer. Since even integers are multiples of 2, the 4500th even integer is 2 multiplied by 4500, which equals 9000.Next, I will calculate the number of digits used when writing all positive even integers up to 9000. I'll break this down into ranges based on the number of digits in the numbers:1. **One-digit even integers**: These are 2, 4, 6, and 8, totaling 4 numbers.2. **Two-digit even integers**: These range from 10 to 98. To find the count, I subtract 10 from 98, divide by 2 (since we're dealing with even numbers), and add 1 to include both endpoints. This gives me 45 numbers.3. **Three-digit even integers**: These range from 100 to 998. Using the same method, I subtract 100 from 998, divide by 2, and add 1, resulting in 450 numbers.4. **Four-digit even integers**: These range from 1000 to 9998. Subtracting 1000 from 9998, dividing by 2, and adding 1 gives me 4500 numbers.Now, I'll calculate the total number of digits for each range:- **One-digit numbers**: 4 numbers × 1 digit = 4 digits- **Two-digit numbers**: 45 numbers × 2 digits = 90 digits- **Three-digit numbers**: 450 numbers × 3 digits = 1350 digits- **Four-digit numbers**: 4500 numbers × 4 digits = 18000 digitsFinally, I'll add up all the digits to find the total number of digits used:4 (one-digit) + 90 (two-digit) + 1350 (three-digit) + 18000 (four-digit) = 19444 digits

💡Okay, so Charles has two dice: one is fair, and the other is biased. The fair die has a 1/6 chance of rolling a six, while the biased die has a 2/3 chance of rolling a six. Charles picks one die at random, so there's a 50% chance he picks either die. He rolls it three times, and we know the first two rolls are sixes. We need to find the probability that the third roll is also a six given this information.Hmm, this sounds like a conditional probability problem. I remember something about Bayes' theorem, which relates the conditional and marginal probabilities of random events. Maybe I can use that here.Let me define the events:- Let F be the event that Charles picked the fair die.- Let B be the event that Charles picked the biased die.- Let S be the event of rolling a six.We know that P(F) = P(B) = 1/2 since he picks one at random.We need to find the probability that the third roll is a six given that the first two were sixes. So, we're looking for P(S | S, S). This can be interpreted as the probability that the third roll is a six given that the first two were sixes.But actually, since the die is the same for all three rolls, maybe I can think of it as updating the probability that Charles is using the fair or biased die after observing two sixes, and then using that to find the probability of the third six.So, first, let's find the probability that Charles is using the fair die given that he rolled two sixes. Similarly, find the probability he's using the biased die given two sixes.Using Bayes' theorem:P(F | S, S) = P(S, S | F) * P(F) / P(S, S)Similarly,P(B | S, S) = P(S, S | B) * P(B) / P(S, S)First, let's compute P(S, S | F). Since the die is fair, each roll is independent, so P(S | F) = 1/6. Therefore, P(S, S | F) = (1/6) * (1/6) = 1/36.Similarly, P(S, S | B) = (2/3) * (2/3) = 4/9.Now, P(S, S) is the total probability of rolling two sixes, regardless of which die is used. So,P(S, S) = P(S, S | F) * P(F) + P(S, S | B) * P(B) = (1/36)*(1/2) + (4/9)*(1/2) = (1/72) + (4/18) = (1/72) + (16/72) = 17/72.Therefore,P(F | S, S) = (1/36 * 1/2) / (17/72) = (1/72) / (17/72) = 1/17.Similarly,P(B | S, S) = (4/9 * 1/2) / (17/72) = (4/18) / (17/72) = (16/72) / (17/72) = 16/17.So, given that Charles rolled two sixes, the probability he's using the fair die is 1/17, and the probability he's using the biased die is 16/17.Now, we need to find the probability that the third roll is a six given that the first two were sixes. This is equivalent to:P(S | S, S) = P(S | F) * P(F | S, S) + P(S | B) * P(B | S, S).We already know P(F | S, S) = 1/17 and P(B | S, S) = 16/17.P(S | F) is 1/6, and P(S | B) is 2/3.Therefore,P(S | S, S) = (1/6)*(1/17) + (2/3)*(16/17) = (1/102) + (32/51).To add these, I need a common denominator. 102 is a multiple of 51, so let's convert 32/51 to 64/102.So,P(S | S, S) = 1/102 + 64/102 = 65/102.Simplify this fraction: 65 and 102 share a common divisor of 13? Let's check.65 ÷ 13 = 5.102 ÷ 13 = 7.846... Hmm, that's not an integer. Wait, maybe 65 and 102 share another common divisor.65 factors: 5, 13.102 factors: 2, 3, 17.No common factors besides 1. So, 65/102 is already in simplest terms.Therefore, the probability is 65/102, so p = 65 and q = 102. Therefore, p + q = 65 + 102 = 167.**Final Answer**boxed{167}

💡Okay, so I have this problem where there are two classes, A and B, that took the same subject exam. Class A has 50 students with an average score of 76 and a variance of 96. Class B has 40 students with an average score of 85 and a variance of 60. I need to find the variance of the scores when combining all 90 students from both classes.Alright, first, I need to recall what variance is. Variance measures how spread out the numbers are. It's the average of the squared differences from the Mean. So, to find the combined variance, I need to consider both classes together.I think the first step is to find the overall average score of all 90 students combined. Since there are different numbers of students in each class, I can't just average the two class averages directly. Instead, I need to calculate a weighted average.So, the formula for the combined average (let's call it μ_combined) would be:μ_combined = (Number of students in A * Average of A + Number of students in B * Average of B) / Total number of studentsPlugging in the numbers:μ_combined = (50 * 76 + 40 * 85) / 90Let me calculate that:First, 50 * 76 = 3800Then, 40 * 85 = 3400Adding those together: 3800 + 3400 = 7200Now, divide by 90: 7200 / 90 = 80So, the combined average score is 80.Okay, now that I have the combined average, I need to find the combined variance. I remember that variance isn't just the average of the variances; it's a bit more involved because it also accounts for how the individual class averages differ from the combined average.The formula for the combined variance (σ²_combined) is:σ²_combined = (n_A * (σ²_A + (μ_A - μ_combined)²) + n_B * (σ²_B + (μ_B - μ_combined)²)) / (n_A + n_B)Where:- n_A and n_B are the number of students in classes A and B- σ²_A and σ²_B are the variances of classes A and B- μ_A and μ_B are the averages of classes A and B- μ_combined is the overall average we calculated earlierLet me plug in the numbers:n_A = 50, σ²_A = 96, μ_A = 76n_B = 40, σ²_B = 60, μ_B = 85μ_combined = 80So,σ²_combined = (50 * (96 + (76 - 80)²) + 40 * (60 + (85 - 80)²)) / 90First, calculate the squared differences:(76 - 80)² = (-4)² = 16(85 - 80)² = 5² = 25Now, plug those back into the equation:σ²_combined = (50 * (96 + 16) + 40 * (60 + 25)) / 90Calculate inside the parentheses:96 + 16 = 11260 + 25 = 85Now:σ²_combined = (50 * 112 + 40 * 85) / 90Calculate the multiplications:50 * 112 = 560040 * 85 = 3400Add them together:5600 + 3400 = 9000Now, divide by 90:9000 / 90 = 100So, the combined variance is 100.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, the combined average was 80, that seems correct.Then, for the variance:For Class A: 50 students, variance 96, average 76. The difference from the combined average is 76 - 80 = -4, squared is 16. So, 96 + 16 = 112. Multiply by 50: 50 * 112 = 5600.For Class B: 40 students, variance 60, average 85. The difference from the combined average is 85 - 80 = 5, squared is 25. So, 60 + 25 = 85. Multiply by 40: 40 * 85 = 3400.Total sum: 5600 + 3400 = 9000.Divide by total number of students: 9000 / 90 = 100.Yes, that seems correct. So, the variance of the combined scores is 100.

💡Okay, so I have this problem about a right triangle with legs measuring 30 inches and 40 inches, and I need to find the length of the hypotenuse. Hmm, right triangles... I remember something about the Pythagorean theorem from math class. Let me try to recall how that works.Alright, the Pythagorean theorem says that in a right-angled triangle, the square of the hypotenuse (which is the side opposite the right angle) is equal to the sum of the squares of the other two sides. So, if I call the legs 'a' and 'b', and the hypotenuse 'c', the formula should be:[ c^2 = a^2 + b^2 ]In this problem, the legs are 30 inches and 40 inches. So, I guess I can plug those values into the formula. Let me write that down:[ c^2 = 30^2 + 40^2 ]Now, I need to calculate 30 squared and 40 squared. Let me do that step by step. First, 30 squared is 30 times 30. Let me multiply that out:30 × 30 = 900Okay, so 30 squared is 900. Now, 40 squared is 40 times 40. Let me calculate that:40 × 40 = 1600Alright, so 40 squared is 1600. Now, I need to add these two results together to find c squared:c² = 900 + 1600Let me add those numbers. 900 plus 1600... Hmm, 900 plus 1000 is 1900, and then plus 600 more would be 2500. So, c squared equals 2500.Now, to find the length of the hypotenuse 'c', I need to take the square root of 2500. I remember that the square root of a number is the value that, when multiplied by itself, gives the original number. So, what number multiplied by itself equals 2500?Let me think. I know that 50 times 50 is 2500 because 5 times 5 is 25, and adding two zeros gives 2500. So, the square root of 2500 is 50.Therefore, the length of the hypotenuse should be 50 inches. Wait, let me double-check my calculations to make sure I didn't make any mistakes. First, 30 squared is 900, and 40 squared is 1600. Adding them together gives 2500. The square root of 2500 is indeed 50. Yeah, that seems correct.I think I've got it right. The hypotenuse is 50 inches long.

💡Okay, so I'm trying to figure out whether "both f(x) and g(x) are even functions" is a sufficient, necessary, both, or neither condition for the product f(x)·g(x) to be an even function. Hmm, let me start by recalling what even functions are.An even function satisfies the condition f(-x) = f(x) for all x in its domain. So, if I have two functions, f and g, both of which are even, then f(-x) = f(x) and g(-x) = g(x). Now, if I take their product, h(x) = f(x)·g(x), I want to check if h(x) is even.Let me compute h(-x):h(-x) = f(-x)·g(-x) = f(x)·g(x) = h(x).So, yes, if both f and g are even, their product is even. That means "both f and g are even" is a sufficient condition for f·g to be even.But is it necessary? In other words, is it possible for f·g to be even even if not both f and g are even? Let me think about that.Suppose one function is even and the other is odd. Let's say f is even and g is odd. Then:h(-x) = f(-x)·g(-x) = f(x)·(-g(x)) = -f(x)·g(x) = -h(x).So, h(x) would be odd in this case, which is not what we want. So, if one is even and the other is odd, the product is odd.What if both functions are odd? Let's see:If f and g are both odd, then:h(-x) = f(-x)·g(-x) = (-f(x))·(-g(x)) = f(x)·g(x) = h(x).So, h(x) is even. Interesting! So, if both f and g are odd, their product is also even.So, this means that there are cases where f·g is even without both f and g being even. Specifically, when both are odd. Therefore, "both f and g are even" is not a necessary condition because f·g can be even even if both f and g are odd.So, putting it all together, "both f and g are even" is a sufficient condition because it guarantees that f·g is even, but it's not necessary because there are other scenarios (like both being odd) where f·g is still even.I should double-check this reasoning. Let me take specific examples.First, let f(x) = x² (which is even) and g(x) = x⁴ (also even). Their product is x⁶, which is even. That checks out.Now, let me take f(x) = x³ (odd) and g(x) = x⁵ (also odd). Their product is x⁸, which is even. So, even though both f and g are odd, their product is even. This confirms that the condition isn't necessary.What if one is even and the other is odd? Let's say f(x) = x² (even) and g(x) = x³ (odd). Their product is x⁵, which is odd. So, that aligns with my earlier conclusion.Therefore, my reasoning seems consistent. The condition that both functions are even is sufficient but not necessary for their product to be even.