💡To determine how many different positive integers can be represented as the difference of two distinct members of the set ( {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ), we can follow these steps:1. **Identify the Range of Possible Differences**: - The smallest possible positive difference is ( 2 - 1 = 1 ). - The largest possible positive difference is ( 10 - 1 = 9 ).2. **Check for Each Difference**: - **Difference of 1**: ( 2 - 1 = 1 ), ( 3 - 2 = 1 ), ..., ( 10 - 9 = 1 ). - **Difference of 2**: ( 3 - 1 = 2 ), ( 4 - 2 = 2 ), ..., ( 10 - 8 = 2 ). - **Difference of 3**: ( 4 - 1 = 3 ), ( 5 - 2 = 3 ), ..., ( 10 - 7 = 3 ). - **Difference of 4**: ( 5 - 1 = 4 ), ( 6 - 2 = 4 ), ..., ( 10 - 6 = 4 ). - **Difference of 5**: ( 6 - 1 = 5 ), ( 7 - 2 = 5 ), ..., ( 10 - 5 = 5 ). - **Difference of 6**: ( 7 - 1 = 6 ), ( 8 - 2 = 6 ), ..., ( 10 - 4 = 6 ). - **Difference of 7**: ( 8 - 1 = 7 ), ( 9 - 2 = 7 ), ( 10 - 3 = 7 ). - **Difference of 8**: ( 9 - 1 = 8 ), ( 10 - 2 = 8 ). - **Difference of 9**: ( 10 - 1 = 9 ).3. **Conclusion**: - Each positive integer from 1 to 9 can be obtained as a difference of two distinct members of the set.Therefore, there are **9** different positive integers that can be represented as the difference of two distinct members of the set.

💡To find the number of elements in the set ( M = {x mid frac{x-5}{x-1} leq 0, x in mathbb{Z}} ), we start by analyzing the inequality ( frac{x-5}{x-1} leq 0 ).First, we note that the expression is undefined when the denominator is zero, so ( x neq 1 ).Next, we determine the critical points where the numerator or denominator is zero: ( x = 5 ) and ( x = 1 ). These points divide the number line into intervals that we can test to determine where the inequality holds.1. **Interval ( x < 1 ):** - Both ( x - 5 ) and ( x - 1 ) are negative. - The product ( (x - 5)(x - 1) ) is positive. - Since we are looking for where the expression is less than or equal to zero, this interval does not satisfy the inequality.2. **Interval ( 1 < x < 5 ):** - ( x - 5 ) is negative and ( x - 1 ) is positive. - The product ( (x - 5)(x - 1) ) is negative. - This interval satisfies the inequality.3. **Interval ( x > 5 ):** - Both ( x - 5 ) and ( x - 1 ) are positive. - The product ( (x - 5)(x - 1) ) is positive. - This interval does not satisfy the inequality.Additionally, we check the critical points:- At ( x = 1 ), the expression is undefined.- At ( x = 5 ), the expression equals zero, which satisfies the inequality.Therefore, the solution to the inequality ( frac{x-5}{x-1} leq 0 ) is ( 1 < x leq 5 ).Since ( x ) must be an integer, the valid integer values are ( 2, 3, 4, ) and ( 5 ).Thus, the set ( M ) contains the elements ( {2, 3, 4, 5} ), and the number of elements in set ( M ) is ( 4 ).

💡First, I need to understand the relationship between speed, time, and distance. The distance the train travels remains constant, so if the speed decreases, the time taken to cover the same distance will increase.Given that the usual time for the journey is 3 hours and the train is 30 minutes late, the total time taken at the reduced speed is 3.5 hours.Let’s denote the usual speed as ( V ) and the reduced speed as ( V' ). The distance can be expressed as ( D = V times 3 ).At the reduced speed, the distance is also ( D = V' times 3.5 ).Since the distances are the same, I can set the two equations equal to each other: ( V times 3 = V' times 3.5 ).To find the fraction of the usual speed, I solve for ( frac{V'}{V} ):[frac{V'}{V} = frac{3}{3.5} = frac{6}{7}]Therefore, the train is moving at ( frac{6}{7} ) of its usual speed.

💡First, I need to determine the number of buffaloes and ducks in the group. Let's denote the number of buffaloes as B and the number of ducks as D.Buffaloes have 4 legs each, and ducks have 2 legs each. The total number of legs in the group is therefore 4B + 2D.Each animal has one head, so the total number of heads is B + D.According to the problem, the number of legs is 24 more than twice the number of heads. This can be expressed as:4B + 2D = 2(B + D) + 24Expanding the right side of the equation:4B + 2D = 2B + 2D + 24Subtracting 2B and 2D from both sides to isolate the variable:2B = 24Dividing both sides by 2 to solve for B:B = 12Therefore, there are 12 buffaloes in the group.

💡First, I need to determine the radii of both the outer and inner circles using their given circumferences. The formula for the circumference of a circle is ( C = 2pi r ), where ( r ) is the radius.For the outer circle with a circumference of ( 24pi ):[24pi = 2pi r_{text{outer}}]Dividing both sides by ( 2pi ):[r_{text{outer}} = frac{24pi}{2pi} = 12]So, the radius of the outer circle is 12, which means ( OB = 12 ).For the inner circle with a circumference of ( 14pi ):[14pi = 2pi r_{text{inner}}]Dividing both sides by ( 2pi ):[r_{text{inner}} = frac{14pi}{2pi} = 7]So, the radius of the inner circle is 7, which means ( OA = 7 ).To find the length of ( AB ), I subtract the radius of the inner circle from the radius of the outer circle:[AB = OB - OA = 12 - 7 = 5]

💡Alright, so I have this sequence problem here. Let me try to understand it step by step.We have a sequence {a_n} where a_1 = 1. The rule for the next term is a bit different depending on whether the current term's index is odd or even. Specifically, if n is odd, then a_{n+1} = a_n + 1, and if n is even, then a_{n+1} = a_n + 2. Part (1) asks me to define a new sequence {b_n} where b_n = a_{2n}. I need to find b_1 and b_2 and then come up with a general formula for {b_n}.Okay, let's start by figuring out what a few terms of the original sequence look like. Since a_1 is given as 1, let's compute a_2, a_3, a_4, and so on, until we can see a pattern.Starting with a_1 = 1.Now, n = 1 is odd, so a_2 = a_1 + 1 = 1 + 1 = 2.Next, n = 2 is even, so a_3 = a_2 + 2 = 2 + 2 = 4.Then, n = 3 is odd, so a_4 = a_3 + 1 = 4 + 1 = 5.n = 4 is even, so a_5 = a_4 + 2 = 5 + 2 = 7.n = 5 is odd, so a_6 = a_5 + 1 = 7 + 1 = 8.n = 6 is even, so a_7 = a_6 + 2 = 8 + 2 = 10.n = 7 is odd, so a_8 = a_7 + 1 = 10 + 1 = 11.n = 8 is even, so a_9 = a_8 + 2 = 11 + 2 = 13.n = 9 is odd, so a_{10} = a_9 + 1 = 13 + 1 = 14.Wait, let me list these out:a_1 = 1a_2 = 2a_3 = 4a_4 = 5a_5 = 7a_6 = 8a_7 = 10a_8 = 11a_9 = 13a_{10} = 14Hmm, interesting. So the sequence alternates between adding 1 and adding 2 depending on whether the current index is odd or even.Now, for part (1), we need to define b_n = a_{2n}. So let's compute b_1 and b_2.b_1 = a_{2*1} = a_2 = 2b_2 = a_{2*2} = a_4 = 5So, b_1 is 2 and b_2 is 5.Now, to find the general formula for {b_n}, which is the sequence of even-indexed terms of {a_n}.Looking at the terms we have:b_1 = 2b_2 = 5Let me compute a few more terms of {a_n} to see if I can spot a pattern in {b_n}.Continuing from where I left off:a_{10} = 14n = 10 is even, so a_{11} = a_{10} + 2 = 14 + 2 = 16n = 11 is odd, so a_{12} = a_{11} + 1 = 16 + 1 = 17n = 12 is even, so a_{13} = a_{12} + 2 = 17 + 2 = 19n = 13 is odd, so a_{14} = a_{13} + 1 = 19 + 1 = 20n = 14 is even, so a_{15} = a_{14} + 2 = 20 + 2 = 22n = 15 is odd, so a_{16} = a_{15} + 1 = 22 + 1 = 23n = 16 is even, so a_{17} = a_{16} + 2 = 23 + 2 = 25n = 17 is odd, so a_{18} = a_{17} + 1 = 25 + 1 = 26n = 18 is even, so a_{19} = a_{18} + 2 = 26 + 2 = 28n = 19 is odd, so a_{20} = a_{19} + 1 = 28 + 1 = 29So, let's list the even-indexed terms:a_2 = 2a_4 = 5a_6 = 8a_8 = 11a_{10} = 14a_{12} = 17a_{14} = 20a_{16} = 23a_{18} = 26a_{20} = 29Looking at these, the sequence {b_n} is: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29,...This seems like an arithmetic sequence where each term increases by 3. Let's check:5 - 2 = 38 - 5 = 311 - 8 = 314 - 11 = 3Yes, so the common difference is 3.Given that, the general formula for an arithmetic sequence is:b_n = b_1 + (n - 1) * dWhere d is the common difference.Here, b_1 = 2 and d = 3.So,b_n = 2 + (n - 1)*3Simplify:b_n = 2 + 3n - 3b_n = 3n - 1So, the general formula for {b_n} is 3n - 1.Wait, let me verify this with the terms we have.For n = 1: 3*1 - 1 = 2 ✔️For n = 2: 3*2 - 1 = 5 ✔️For n = 3: 3*3 - 1 = 8 ✔️For n = 4: 3*4 - 1 = 11 ✔️Yes, this seems correct.So, part (1) is done. b_1 = 2, b_2 = 5, and the general formula is b_n = 3n - 1.Now, moving on to part (2): Find the sum of the first 20 terms of the sequence {a_n}.So, I need to compute S = a_1 + a_2 + a_3 + ... + a_{20}.Given that, let's see if I can find a pattern or formula for a_n so that I can compute the sum without having to calculate each term individually.Looking back at the terms we have:a_1 = 1a_2 = 2a_3 = 4a_4 = 5a_5 = 7a_6 = 8a_7 = 10a_8 = 11a_9 = 13a_{10} = 14a_{11} = 16a_{12} = 17a_{13} = 19a_{14} = 20a_{15} = 22a_{16} = 23a_{17} = 25a_{18} = 26a_{19} = 28a_{20} = 29Looking at these, it seems that the odd-indexed terms and even-indexed terms form their own sequences.From part (1), we have the formula for the even-indexed terms: b_n = a_{2n} = 3n - 1.What about the odd-indexed terms? Let's see.Looking at a_1, a_3, a_5, a_7, etc.:a_1 = 1a_3 = 4a_5 = 7a_7 = 10a_9 = 13a_{11} = 16a_{13} = 19a_{15} = 22a_{17} = 25a_{19} = 28This also seems like an arithmetic sequence. Let's check the differences:4 - 1 = 37 - 4 = 310 - 7 = 313 - 10 = 316 - 13 = 319 - 16 = 322 - 19 = 325 - 22 = 328 - 25 = 3Yes, so the odd-indexed terms also form an arithmetic sequence with a common difference of 3.What's the first term? a_1 = 1.So, the general formula for the odd-indexed terms, let's denote it as c_n = a_{2n - 1}.Then, c_n = 1 + (n - 1)*3 = 3n - 2.Let me verify:For n = 1: 3*1 - 2 = 1 ✔️For n = 2: 3*2 - 2 = 4 ✔️For n = 3: 3*3 - 2 = 7 ✔️Yes, that works.So, now we have:For odd indices: a_{2n - 1} = 3n - 2For even indices: a_{2n} = 3n - 1Therefore, the first 20 terms consist of 10 odd-indexed terms and 10 even-indexed terms.So, the sum S = sum_{n=1}^{20} a_n = sum_{n=1}^{10} a_{2n - 1} + sum_{n=1}^{10} a_{2n}Which is equal to sum_{n=1}^{10} (3n - 2) + sum_{n=1}^{10} (3n - 1)Let me compute each sum separately.First, sum_{n=1}^{10} (3n - 2):This is an arithmetic series where each term is 3n - 2.The first term when n=1: 3*1 - 2 = 1The last term when n=10: 3*10 - 2 = 30 - 2 = 28Number of terms = 10Sum = (number of terms)/2 * (first term + last term) = 10/2 * (1 + 28) = 5 * 29 = 145Similarly, sum_{n=1}^{10} (3n - 1):First term when n=1: 3*1 - 1 = 2Last term when n=10: 3*10 - 1 = 30 - 1 = 29Number of terms = 10Sum = 10/2 * (2 + 29) = 5 * 31 = 155Therefore, total sum S = 145 + 155 = 300Wait, let me double-check these calculations.For the odd terms:Sum = 1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 + 25 + 28Let me add them step by step:1 + 4 = 55 + 7 = 1212 + 10 = 2222 + 13 = 3535 + 16 = 5151 + 19 = 7070 + 22 = 9292 + 25 = 117117 + 28 = 145Yes, that's correct.For the even terms:2 + 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29Adding step by step:2 + 5 = 77 + 8 = 1515 + 11 = 2626 + 14 = 4040 + 17 = 5757 + 20 = 7777 + 23 = 100100 + 26 = 126126 + 29 = 155Yes, that's correct too.So, total sum is indeed 145 + 155 = 300.Therefore, the sum of the first 20 terms is 300.Alternatively, I could have noticed that each pair of terms (a_{2n-1}, a_{2n}) adds up to (3n - 2) + (3n - 1) = 6n - 3.Then, the sum over n from 1 to 10 would be sum_{n=1}^{10} (6n - 3) = 6*sum(n) - 3*10.Sum(n) from 1 to 10 is (10*11)/2 = 55.So, 6*55 = 3303*10 = 30Thus, total sum = 330 - 30 = 300.Same result.So, either way, the sum is 300.I think that's solid.**Final Answer**The sum of the first 20 terms of the sequence is boxed{300}.