💡Okay, so I have this congruence to solve: 7(5x - 3) ≡ 35 mod 10, and I need to find all positive integer solutions x that are less than or equal to 30, then sum them up. Hmm, let me try to figure this out step by step.First, I remember that when dealing with congruences, I can simplify them by breaking them down. So, let me start by expanding the left side of the congruence. 7 times (5x - 3) is... let me calculate that. 7*5x is 35x, and 7*(-3) is -21. So, the congruence becomes 35x - 21 ≡ 35 mod 10.Now, I can simplify this modulo 10. Let me subtract 35 from both sides to get 35x - 21 - 35 ≡ 0 mod 10. That simplifies to 35x - 56 ≡ 0 mod 10. Hmm, but 35x is the same as 5x mod 10 because 35 divided by 10 leaves a remainder of 5. Similarly, 56 mod 10 is 6. So, substituting that in, I get 5x - 6 ≡ 0 mod 10.Wait, that doesn't seem right. Let me double-check. If I have 35x - 21 ≡ 35 mod 10, then subtracting 35 from both sides gives 35x - 56 ≡ 0 mod 10. Now, 35x mod 10 is indeed 5x, and 56 mod 10 is 6. So, 5x - 6 ≡ 0 mod 10. That seems correct.So, 5x - 6 ≡ 0 mod 10. Let me solve for x. Adding 6 to both sides gives 5x ≡ 6 mod 10. Now, I need to find x such that when multiplied by 5, it leaves a remainder of 6 modulo 10.Hmm, 5x ≡ 6 mod 10. Let me think about possible x values. If x is 0, 5*0=0≡0 mod10. x=1, 5*1=5≡5 mod10. x=2, 5*2=10≡0 mod10. x=3, 5*3=15≡5 mod10. x=4, 5*4=20≡0 mod10. x=5, 5*5=25≡5 mod10. x=6, 5*6=30≡0 mod10. x=7, 5*7=35≡5 mod10. x=8, 5*8=40≡0 mod10. x=9, 5*9=45≡5 mod10. x=10, 5*10=50≡0 mod10.Wait, I'm not getting 6 anywhere. So, does that mean there's no solution? That can't be right because the problem is asking for solutions. Maybe I made a mistake earlier.Let me go back. The original congruence was 7(5x - 3) ≡ 35 mod 10. Maybe instead of expanding, I can simplify 7(5x - 3) first. Let's see, 7 and 10 are not coprime, so maybe I can divide both sides by a common factor.Wait, 35 mod 10 is 5, right? So, 7(5x - 3) ≡ 5 mod 10. Let me write that as 7*(5x - 3) ≡ 5 mod 10.Now, 7 and 10 are coprime? No, their GCD is 1, so they are coprime. Wait, 7 and 10 have GCD 1, so 7 has an inverse modulo 10. Let me find the inverse of 7 mod 10. 7*3=21≡1 mod10, so the inverse is 3.So, multiplying both sides by 3: 3*7*(5x - 3) ≡ 3*5 mod10. That simplifies to (21)*(5x - 3) ≡ 15 mod10. But 21≡1 mod10, so this becomes 1*(5x - 3) ≡ 15 mod10. Therefore, 5x - 3 ≡ 15 mod10.Now, 15 mod10 is 5, so 5x - 3 ≡5 mod10. Adding 3 to both sides: 5x ≡8 mod10.Wait, earlier I had 5x≡6 mod10, but now I have 5x≡8 mod10. Hmm, which one is correct? Let me check my steps.Original congruence: 7(5x - 3) ≡35 mod10. 35 mod10 is 5, so 7(5x - 3)≡5 mod10. Then, since 7 and 10 are coprime, multiply both sides by 3 (the inverse of 7 mod10): 3*7*(5x -3)≡3*5 mod10. 3*7=21≡1 mod10, so 1*(5x -3)≡15 mod10. 15 mod10 is 5, so 5x -3≡5 mod10. Adding 3: 5x≡8 mod10.So, 5x≡8 mod10. Now, let's solve for x. 5x≡8 mod10. Let's see, 5x mod10 can be 0 or 5, depending on x. If x is even, 5x≡0 mod10; if x is odd, 5x≡5 mod10. So, 5x can never be 8 mod10. That means there's no solution? But the problem says to find all positive integer solutions less than or equal to 30. That doesn't make sense.Wait, maybe I made a mistake earlier. Let me try another approach. Starting from the beginning: 7(5x -3)≡35 mod10.Let me compute 7(5x -3) mod10. 7*5x=35x≡5x mod10, and 7*(-3)=-21≡-1 mod10. So, 5x -1≡35 mod10. But 35 mod10 is 5, so 5x -1≡5 mod10. Adding 1 to both sides: 5x≡6 mod10.Wait, now I'm back to 5x≡6 mod10. Earlier, I thought there was no solution because 5x mod10 is either 0 or 5. But 6 isn't 0 or 5, so no solution? But the problem says to find solutions. Maybe I'm missing something.Wait, perhaps I should consider that 5x≡6 mod10 can be rewritten as 5x -6≡0 mod10. So, 5x -6 is divisible by 10. Let me write that as 5x = 10k +6, where k is an integer. Then, x=(10k +6)/5=2k + 6/5. Hmm, but 6/5 is not an integer, so x would not be an integer unless 10k +6 is divisible by 5. But 10k is divisible by 5, and 6 mod5 is 1, so 10k +6≡1 mod5. Therefore, 10k +6 is not divisible by5, so x is not an integer. Therefore, no solution exists.But the problem says to find all positive integer solutions less than or equal to30. So, is there a mistake in my calculations?Wait, let me try another approach. Maybe I should simplify the original congruence differently. Starting again: 7(5x -3)≡35 mod10.Let me compute 7(5x -3) mod10. 7*5x=35x≡5x mod10, and 7*(-3)=-21≡-1 mod10. So, 5x -1≡35 mod10. 35 mod10 is 5, so 5x -1≡5 mod10. Adding 1: 5x≡6 mod10.Again, same result. So, 5x≡6 mod10. As before, 5x mod10 can only be 0 or5, so 6 is not possible. Therefore, no solution exists.But the problem says to find all positive integer solutions less than or equal to30. Maybe I'm misunderstanding the problem. Let me check the original problem again.It says: What is the sum of all positive integer solutions to the congruence 7(5x -3)≡35 mod10, where the solutions are less than or equal to30.Wait, maybe I should consider that 35 mod10 is 5, so 7(5x -3)≡5 mod10. Let me write that as 7*(5x -3)≡5 mod10. Since 7 and10 are coprime, I can multiply both sides by the inverse of7 mod10, which is3, as before. So, 3*7*(5x -3)≡3*5 mod10. 21*(5x -3)≡15 mod10. 21≡1 mod10, so 1*(5x -3)≡15 mod10. Therefore, 5x -3≡15 mod10. 15 mod10 is5, so 5x -3≡5 mod10. Adding3:5x≡8 mod10.Again, same result. 5x≡8 mod10. As before, no solution because 5x mod10 is 0 or5, not8.Wait, maybe I should check if 5 and10 are coprime. GCD(5,10)=5, so they are not coprime. Therefore, for the congruence5x≡8 mod10 to have solutions, 8 must be divisible by GCD(5,10)=5. But 8 is not divisible by5, so no solution exists.Therefore, there are no positive integer solutions to the congruence7(5x -3)≡35 mod10. Hence, the sum is0.But the problem says to find the sum of all positive integer solutions less than or equal to30. If there are no solutions, the sum is0. But maybe I made a mistake somewhere.Wait, let me try plugging in some values forx to see if any satisfy the congruence. Let's tryx=1:7(5*1 -3)=7(5-3)=7*2=14. 14 mod10=4. 4≡35 mod10? 35 mod10=5. 4≡5 mod10? No.x=2:7(10-3)=7*7=49. 49 mod10=9. 9≡5? No.x=3:7(15-3)=7*12=84. 84 mod10=4. 4≡5? No.x=4:7(20-3)=7*17=119. 119 mod10=9. 9≡5? No.x=5:7(25-3)=7*22=154. 154 mod10=4. 4≡5? No.x=6:7(30-3)=7*27=189. 189 mod10=9. 9≡5? No.x=7:7(35-3)=7*32=224. 224 mod10=4. 4≡5? No.x=8:7(40-3)=7*37=259. 259 mod10=9. 9≡5? No.x=9:7(45-3)=7*42=294. 294 mod10=4. 4≡5? No.x=10:7(50-3)=7*47=329. 329 mod10=9. 9≡5? No.Hmm, none of these x values satisfy the congruence. So, it seems there are no solutions. Therefore, the sum is0.But the problem is asking for the sum of all positive integer solutions less than or equal to30. If there are no solutions, the sum is0. But maybe I made a mistake in simplifying the congruence.Wait, let me try another approach. Let's go back to the original congruence:7(5x -3)≡35 mod10.Let me compute both sides modulo10. 7(5x -3) mod10. 7*5x=35x≡5x mod10, and 7*(-3)=-21≡-1 mod10. So, 5x -1≡35 mod10. 35 mod10=5, so 5x -1≡5 mod10. Adding1:5x≡6 mod10.Again, same result. So, 5x≡6 mod10. As before, no solution because 5x mod10 is either0 or5, not6.Therefore, there are no solutions. Hence, the sum is0.But the problem is from a math competition, and usually, such problems have solutions. Maybe I misread the problem. Let me check again.It says: What is the sum of all positive integer solutions to the congruence7(5x -3)≡35 mod10, where the solutions are less than or equal to30.Wait, maybe I should consider that 35 mod10 is5, so 7(5x -3)≡5 mod10. Let me write that as7*(5x -3)≡5 mod10.Since7 and10 are coprime, I can multiply both sides by the inverse of7 mod10, which is3. So, 3*7*(5x -3)≡3*5 mod10. 21*(5x -3)≡15 mod10. 21≡1 mod10, so 1*(5x -3)≡15 mod10. Therefore,5x -3≡15 mod10. 15 mod10=5, so5x -3≡5 mod10. Adding3:5x≡8 mod10.Again, same result. 5x≡8 mod10. As before, no solution.Wait, maybe I should consider that the modulus is10, so the solutions repeat every10 numbers. But since there's no solution, there are no solutions in any modulus.Therefore, the sum is0.But the problem is asking for the sum, so maybe the answer is0.Wait, but in the initial problem, the user wrote the same problem and got an answer of225. Maybe I did something wrong.Wait, let me check the initial problem again. The user wrote: What is the sum of all positive integer solutions to the congruence7(5x -3)≡35 mod10, where the solutions are less than or equal to30.In the initial problem, the user's assistant got an answer of225 by assuming x≡1 mod2, leading to solutions1,3,5,...,29, summing to225.But according to my calculations, there are no solutions because5x≡6 mod10 has no solution. So, why did the assistant get225?Wait, maybe the assistant made a mistake. Let me see.In the initial problem, the assistant wrote:7(5x -3)≡35 mod10.35x -21≡35 mod10.35x -21≡5 mod10.35x≡26 mod10.5x≡6 mod10.x≡1.2 mod2.Then, the assistant realized that x≡1.2 mod2 doesn't make sense, so they tried dividing by7:5x -1≡5 mod10.5x≡6 mod10.x≡6/5 mod2.Again, non-integer.Then, the assistant tried another approach:7(5x -3)≡35 mod10.35x -21≡35 mod10.5x -1≡5 mod2.5x≡6 mod2.x≡1.2 mod2.Again, non-integer.Then, the assistant assumed x≡1 mod2, leading to solutions1,3,5,...,29, summing to225.But according to my calculations, there are no solutions because5x≡6 mod10 has no solution.So, the assistant made a mistake by assuming x≡1 mod2, which is not correct.Therefore, the correct answer is that there are no solutions, so the sum is0.But the problem is asking for the sum, so maybe the answer is0.Wait, but in the initial problem, the assistant got225, which is incorrect. So, the correct answer is0.But let me double-check. Maybe I'm missing something.Wait, let me try to solve5x≡6 mod10.5x≡6 mod10.Let me write this as5x -6=10k, where k is integer.So,5x=10k +6.Divide both sides by5: x=2k +6/5.But6/5 is not an integer, so x is not an integer. Therefore, no solution.Hence, the sum is0.Therefore, the correct answer is0.

💡Okay, so I have this problem here about a function, y equals x squared minus 2x plus 5. It says that on the interval [0, m], the function has a maximum value of 5 and a minimum value of 4. I need to figure out what the range of the real number m is, and the options are given as A through D.First, I remember that for quadratic functions, the graph is a parabola. Since the coefficient of x squared is positive (which is 1 in this case), the parabola opens upwards. That means the vertex of the parabola is the minimum point. So, the function should have a minimum value at its vertex and then it increases as we move away from the vertex in both directions.Let me find the vertex of this function. The general form of a quadratic is ax² + bx + c, and the x-coordinate of the vertex is given by -b/(2a). Here, a is 1 and b is -2. So, plugging in, the x-coordinate is -(-2)/(2*1) which is 2/2, so 1. Therefore, the vertex is at x = 1.To find the y-coordinate of the vertex, I plug x = 1 back into the function: y = (1)^2 - 2*(1) + 5 = 1 - 2 + 5 = 4. So, the minimum value of the function is 4 at x = 1.Now, the problem says that on the interval [0, m], the function has a maximum value of 5 and a minimum value of 4. Since the minimum is 4, which occurs at x = 1, that means our interval [0, m] must include x = 1. So, m has to be at least 1, right? Because if m were less than 1, the interval [0, m] wouldn't include the vertex, and the minimum might not be 4.But wait, let me think again. If m is less than 1, say m = 0.5, then the function on [0, 0.5] would be decreasing because the parabola opens upwards and the vertex is at x = 1. So, the minimum would be at x = 0.5, but the function at x = 0 is y = 0 - 0 + 5 = 5. So, on [0, 0.5], the function starts at 5 when x=0 and decreases to y = (0.5)^2 - 2*(0.5) + 5 = 0.25 - 1 + 5 = 4.25. So, the minimum would be 4.25, not 4. Therefore, to have the minimum value of 4, the interval must include x = 1.So, m must be at least 1. If m is exactly 1, then the interval is [0,1]. On this interval, the function starts at y=5 when x=0, decreases to y=4 at x=1. So, the maximum is 5 and the minimum is 4, which fits the problem's condition.Now, what happens if m is greater than 1? Let's say m is 2. Then, the interval is [0,2]. On this interval, the function starts at 5 when x=0, decreases to 4 at x=1, and then increases again. Let's compute y at x=2: y = (2)^2 - 2*(2) + 5 = 4 - 4 + 5 = 5. So, at x=2, it's back to 5. So, on [0,2], the function goes from 5 down to 4 and back up to 5. So, the maximum is still 5 and the minimum is 4. So, m can be 2.What if m is more than 2? Let's say m is 3. Then, the interval is [0,3]. On this interval, the function starts at 5, goes down to 4 at x=1, then increases beyond 5 as x increases beyond 2. So, at x=3, y = 9 - 6 + 5 = 8. So, the maximum value would be 8, which is higher than 5. But the problem states that the maximum is 5. Therefore, m cannot be more than 2 because beyond x=2, the function exceeds 5.Wait, let me check at x=2, y is 5, and beyond that, it increases. So, if m is greater than 2, the function will have a maximum higher than 5, which contradicts the problem's condition. Therefore, m cannot be greater than 2.So, putting it all together, m has to be at least 1 to include the minimum at x=1, and it can't be more than 2 because beyond that, the maximum exceeds 5. Therefore, m must be between 1 and 2, inclusive.Let me just verify this with another value. If m is 1.5, then the interval is [0,1.5]. The function starts at 5, goes down to 4 at x=1, then increases again. At x=1.5, y = (1.5)^2 - 2*(1.5) + 5 = 2.25 - 3 + 5 = 4.25. So, the maximum is still 5 (at x=0 and x=1.5, but wait, at x=1.5, y is 4.25, which is less than 5. So, the maximum is still 5 at x=0. So, that's okay. The minimum is 4 at x=1. So, that works.If m is 2, as I saw earlier, the function goes back up to 5 at x=2, so the maximum is still 5, and the minimum is 4. So, that's acceptable.If m is less than 1, say 0.5, as I checked before, the minimum is 4.25, which is higher than 4, so that's not acceptable.Therefore, the range of m is from 1 to 2, inclusive. So, the correct option is D: [1, 2].I think that makes sense. I don't see any mistakes in my reasoning. The key points are identifying the vertex as the minimum, ensuring that the interval includes the vertex, and making sure that beyond a certain point, the function doesn't exceed the given maximum. So, yeah, D is the right answer.

💡Alright, so I have this problem here: the function ( f(x) = x^2 + 2x + 3 ) has a maximum value of 3 and a minimum value of 2 on the interval ([m, 0]). I need to find the range of possible values for ( m ). The options are A: [1,2], B: ((-infty, -1]), C: [-1,0], and D: [-2,-1]. Hmm, okay, let me try to figure this out step by step.First, I remember that quadratic functions have either a maximum or a minimum value depending on the coefficient of ( x^2 ). In this case, the coefficient is positive (1), so the parabola opens upwards, meaning it has a minimum point. But the problem is talking about both a maximum and a minimum on a specific interval. So, I guess the interval [m, 0] must be such that the function reaches both its maximum and minimum within that interval.Maybe I should start by finding the vertex of the parabola because that's where the minimum occurs. The vertex form of a quadratic function ( ax^2 + bx + c ) is at ( x = -frac{b}{2a} ). Here, ( a = 1 ) and ( b = 2 ), so the vertex is at ( x = -frac{2}{2*1} = -1 ). So, the minimum value of the function is at ( x = -1 ). Let me calculate that: ( f(-1) = (-1)^2 + 2*(-1) + 3 = 1 - 2 + 3 = 2 ). Okay, so the minimum value is indeed 2 at ( x = -1 ).Now, the maximum value on the interval [m, 0] is given as 3. Since the parabola opens upwards, the maximum on an interval will occur at one of the endpoints. So, let's check the endpoints: at ( x = 0 ) and ( x = m ).Calculating ( f(0) ): ( 0^2 + 2*0 + 3 = 3 ). So, at ( x = 0 ), the function is 3. That's our maximum. Now, what about ( x = m )? We need ( f(m) ) to be less than or equal to 3 because the maximum is at 0. So, ( f(m) leq 3 ).Let me write that out: ( m^2 + 2m + 3 leq 3 ). Simplifying that inequality: ( m^2 + 2m leq 0 ). Factoring: ( m(m + 2) leq 0 ). So, this inequality holds when ( m ) is between -2 and 0, inclusive. Because the product of two numbers is less than or equal to zero when one is non-positive and the other is non-negative.But wait, the interval is [m, 0], so ( m ) has to be less than or equal to 0. From the inequality, ( m ) can be between -2 and 0. But we also know that the minimum value of 2 occurs at ( x = -1 ). So, for the function to attain its minimum at ( x = -1 ), the interval [m, 0] must include ( x = -1 ). That means ( m ) must be less than or equal to -1 because if ( m ) were greater than -1, say -0.5, then the interval would be [-0.5, 0], which doesn't include -1, and the minimum wouldn't be attained there.So, combining these two pieces of information: ( m ) must be between -2 and 0, but also ( m ) must be less than or equal to -1 to include the point where the minimum occurs. Therefore, ( m ) must be between -2 and -1, inclusive.Let me double-check that. If ( m = -2 ), then the interval is [-2, 0]. The function at -2 is ( (-2)^2 + 2*(-2) + 3 = 4 - 4 + 3 = 3 ), which is the maximum. The function at -1 is 2, which is the minimum. So, that works. If ( m = -1 ), the interval is [-1, 0]. The function at -1 is 2, and at 0 is 3. So, that also works. If ( m ) is between -2 and -1, say -1.5, then the interval is [-1.5, 0]. The function at -1.5 is ( (-1.5)^2 + 2*(-1.5) + 3 = 2.25 - 3 + 3 = 2.25 ), which is less than 3 but more than 2. The function at -1 is still 2, which is the minimum, and at 0 is 3, the maximum. So, that also works.What if ( m ) is less than -2? Let's say ( m = -3 ). Then the interval is [-3, 0]. The function at -3 is ( (-3)^2 + 2*(-3) + 3 = 9 - 6 + 3 = 6 ), which is greater than 3. But the maximum is supposed to be 3, so that doesn't work. Therefore, ( m ) can't be less than -2.What if ( m ) is greater than -1? Let's say ( m = -0.5 ). Then the interval is [-0.5, 0]. The function at -0.5 is ( (-0.5)^2 + 2*(-0.5) + 3 = 0.25 - 1 + 3 = 2.25 ). The maximum is at 0, which is 3, and the minimum is at -0.5, which is 2.25. But the problem states the minimum should be 2, which isn't achieved here because -1 isn't in the interval. So, that doesn't work either.Therefore, the only valid range for ( m ) is between -2 and -1, inclusive. So, the correct answer should be D: [-2,-1].**Final Answer**boxed{D}

💡Okay, so I have this problem here about a positive sequence {a_n}. The sum of its first n terms, which is S_n, satisfies the equation 10S_n = a_n² + 5a_n + 6. Also, it's given that a₁, a₃, a₁₅ form a geometric sequence. I need to find the general term a_n of this sequence.Hmm, let me start by understanding what's given. The sum S_n is related to the nth term a_n through this quadratic equation. That seems a bit unusual because normally, the sum of terms is linear in the terms, but here it's quadratic. So, maybe I can express S_n in terms of a_n and then find a recursive relation or something.First, let me write down the given equation:10S_n = a_n² + 5a_n + 6.I know that S_n is the sum of the first n terms, so S_n = a₁ + a₂ + ... + a_n. Similarly, S_{n-1} would be the sum of the first n-1 terms. So, if I subtract S_{n-1} from S_n, I get a_n:a_n = S_n - S_{n-1}.But in the given equation, I have 10S_n = a_n² + 5a_n + 6. Maybe I can write a similar equation for S_{n-1}:10S_{n-1} = a_{n-1}² + 5a_{n-1} + 6.Now, if I subtract these two equations, I can get an expression involving a_n and a_{n-1}:10S_n - 10S_{n-1} = (a_n² + 5a_n + 6) - (a_{n-1}² + 5a_{n-1} + 6).Simplifying the left side: 10(S_n - S_{n-1}) = 10a_n.On the right side: a_n² - a_{n-1}² + 5a_n - 5a_{n-1}.So, putting it all together:10a_n = (a_n² - a_{n-1}²) + 5(a_n - a_{n-1}).Let me factor the right side. The term a_n² - a_{n-1}² is a difference of squares, so it factors into (a_n - a_{n-1})(a_n + a_{n-1}). Similarly, 5(a_n - a_{n-1}) is just 5 times the difference.So, the equation becomes:10a_n = (a_n - a_{n-1})(a_n + a_{n-1}) + 5(a_n - a_{n-1}).I can factor out (a_n - a_{n-1}) from both terms on the right:10a_n = (a_n - a_{n-1})(a_n + a_{n-1} + 5).Hmm, so now I have:10a_n = (a_n - a_{n-1})(a_n + a_{n-1} + 5).This seems a bit complicated, but maybe I can rearrange it. Let me bring everything to one side:10a_n - (a_n - a_{n-1})(a_n + a_{n-1} + 5) = 0.Expanding the product:10a_n - [a_n(a_n + a_{n-1} + 5) - a_{n-1}(a_n + a_{n-1} + 5)] = 0.Wait, that might not be the most straightforward way. Maybe instead, let me consider moving 10a_n to the other side:(a_n - a_{n-1})(a_n + a_{n-1} + 5) = 10a_n.Hmm, not sure if that helps. Maybe I can divide both sides by (a_n - a_{n-1}), assuming it's not zero. But since the sequence is positive, and the terms are increasing or decreasing? Wait, I don't know yet.Alternatively, maybe I can let d_n = a_n - a_{n-1}, so the difference between consecutive terms. Then, the equation becomes:10a_n = d_n(a_n + a_{n-1} + 5).But I'm not sure if that substitution helps immediately.Wait, maybe I can rearrange the equation:10a_n = (a_n - a_{n-1})(a_n + a_{n-1} + 5).Let me divide both sides by (a_n + a_{n-1} + 5):10a_n / (a_n + a_{n-1} + 5) = a_n - a_{n-1}.Hmm, that might be useful. Let me denote this as:a_n - a_{n-1} = 10a_n / (a_n + a_{n-1} + 5).Let me rearrange this:(a_n - a_{n-1})(a_n + a_{n-1} + 5) = 10a_n.Wait, that's the same as before. Maybe I can make a substitution. Let me set b_n = a_n + a_{n-1} + 5. Then, the equation becomes:(a_n - a_{n-1}) * b_n = 10a_n.But I don't know if that helps. Maybe another approach.Alternatively, let me consider that the equation is quadratic in a_n. Let me rearrange the equation:10a_n = (a_n - a_{n-1})(a_n + a_{n-1} + 5).Expanding the right side:10a_n = a_n² + a_n a_{n-1} + 5a_n - a_{n-1}a_n - a_{n-1}² - 5a_{n-1}.Simplify:10a_n = a_n² + 5a_n - a_{n-1}² - 5a_{n-1}.Bring all terms to one side:0 = a_n² + 5a_n - a_{n-1}² - 5a_{n-1} - 10a_n.Simplify:0 = a_n² - 5a_n - a_{n-1}² - 5a_{n-1}.Hmm, that's:a_n² - 5a_n = a_{n-1}² + 5a_{n-1}.So, a_n² - 5a_n - a_{n-1}² - 5a_{n-1} = 0.Let me factor this as:(a_n² - a_{n-1}²) - 5(a_n + a_{n-1}) = 0.Factor the difference of squares:(a_n - a_{n-1})(a_n + a_{n-1}) - 5(a_n + a_{n-1}) = 0.Factor out (a_n + a_{n-1}):(a_n + a_{n-1})(a_n - a_{n-1} - 5) = 0.Ah, that's a key step. So, either (a_n + a_{n-1}) = 0 or (a_n - a_{n-1} - 5) = 0.But since the sequence is positive, a_n + a_{n-1} is always positive, so it can't be zero. Therefore, the other factor must be zero:a_n - a_{n-1} - 5 = 0 ⇒ a_n = a_{n-1} + 5.So, this tells me that the sequence {a_n} is an arithmetic sequence with common difference 5. That's a big revelation.So, if it's an arithmetic sequence with common difference 5, then a_n = a₁ + (n - 1)*5.Now, I need to find a₁. For that, I can use the initial condition given by the equation when n=1.When n=1, S₁ = a₁. Plugging into the given equation:10S₁ = a₁² + 5a₁ + 6 ⇒ 10a₁ = a₁² + 5a₁ + 6.Let me solve this quadratic equation:a₁² + 5a₁ + 6 - 10a₁ = 0 ⇒ a₁² - 5a₁ + 6 = 0.Factoring:(a₁ - 2)(a₁ - 3) = 0 ⇒ a₁ = 2 or a₁ = 3.So, possible values for a₁ are 2 or 3. Now, I need to determine which one is correct.But wait, there's another condition: a₁, a₃, a₁₅ form a geometric sequence. So, let's compute a₃ and a₁₅ for both cases and check if they form a geometric sequence.Case 1: a₁ = 2.Since it's an arithmetic sequence with d=5, a₃ = a₁ + 2d = 2 + 10 = 12.Similarly, a₁₅ = a₁ + 14d = 2 + 70 = 72.So, the terms are 2, 12, 72. Let's check if they form a geometric sequence.In a geometric sequence, the ratio between consecutive terms should be constant.So, 12 / 2 = 6, and 72 / 12 = 6. So, yes, the common ratio is 6. Therefore, this case satisfies the condition.Case 2: a₁ = 3.Then, a₃ = 3 + 10 = 13.a₁₅ = 3 + 70 = 73.So, the terms are 3, 13, 73. Let's check the ratios:13 / 3 ≈ 4.333..., and 73 / 13 ≈ 5.615... These are not equal, so they don't form a geometric sequence.Therefore, a₁ cannot be 3. So, the correct value is a₁ = 2.Thus, the general term is a_n = a₁ + (n - 1)*5 = 2 + 5(n - 1) = 5n - 3.Let me double-check this result.First, check the initial condition for n=1:a₁ = 5*1 - 3 = 2, which is correct.Compute S_n for n=1: S₁ = 2.10S₁ = 20. On the other side, a₁² + 5a₁ + 6 = 4 + 10 + 6 = 20. So, it matches.Now, check for n=2:a₂ = 5*2 - 3 = 7.S₂ = a₁ + a₂ = 2 + 7 = 9.10S₂ = 90.a₂² + 5a₂ + 6 = 49 + 35 + 6 = 90. Correct.Similarly, for n=3:a₃ = 12, S₃ = 2 + 7 + 12 = 21.10S₃ = 210.a₃² + 5a₃ + 6 = 144 + 60 + 6 = 210. Correct.Good, so the formula seems to hold.Also, checking the geometric sequence condition:a₁ = 2, a₃ = 12, a₁₅ = 72.12 / 2 = 6, 72 / 12 = 6. So, it's a geometric sequence with ratio 6. Perfect.Therefore, the general term is a_n = 5n - 3.

💡Okay, so I have this problem about a right triangle ABC with hypotenuse AB. One of the legs, AC, is 15 units long. There's an altitude from point C to the hypotenuse AB, and this altitude divides AB into two segments: AH and HB. They tell me that HB is 16 units. I need to find the area of triangle ABC.Alright, let's start by drawing a diagram in my mind. Triangle ABC is right-angled, so let's assume the right angle is at point C. That means AC and BC are the legs, and AB is the hypotenuse. The altitude from C to AB will meet AB at some point, let's call it H. So, AH and HB are the two segments of AB created by this altitude.I remember that in a right triangle, the altitude to the hypotenuse has some special properties. Specifically, the length of the altitude is the geometric mean of the lengths of the two segments it creates on the hypotenuse. Also, each leg of the triangle is the geometric mean of the hypotenuse and the adjacent segment.Let me write down what I know:- AC = 15 (one leg)- HB = 16 (one segment of the hypotenuse)- Let AH = x (the other segment of the hypotenuse)- So, AB = AH + HB = x + 16Since ABC is a right triangle, by the Pythagorean theorem:AB² = AC² + BC²But I don't know BC yet. Maybe I can find BC using the properties of the altitude.I recall that in a right triangle, the altitude to the hypotenuse relates the segments AH and HB. Specifically, the length of the altitude CH is equal to the square root of (AH * HB). So:CH = √(AH * HB) = √(x * 16)Also, the area of triangle ABC can be expressed in two ways: one using the legs AC and BC, and another using the hypotenuse AB and the altitude CH.So, area = (1/2) * AC * BC = (1/2) * AB * CHSince I need to find the area, maybe I can express BC in terms of x and then solve for x.But wait, let's see if I can find x first. Maybe using the geometric mean property.Another property I remember is that each leg is the geometric mean of the hypotenuse and the adjacent segment. So:AC² = AH * ABSimilarly,BC² = HB * ABSo, let's use the first one:AC² = AH * AB15² = x * (x + 16)225 = x² + 16xSo, x² + 16x - 225 = 0This is a quadratic equation in terms of x. Let's solve it.Using the quadratic formula:x = [-b ± √(b² - 4ac)] / 2aHere, a = 1, b = 16, c = -225Discriminant D = 16² - 4*1*(-225) = 256 + 900 = 1156√D = √1156 = 34So,x = [-16 ± 34] / 2We have two solutions:x = (-16 + 34)/2 = 18/2 = 9x = (-16 - 34)/2 = -50/2 = -25Since length can't be negative, x = 9So, AH = 9, HB = 16, so AB = 9 + 16 = 25Now, let's find BC using the other geometric mean property:BC² = HB * AB = 16 * 25 = 400So, BC = √400 = 20Now, we can find the area using the legs:Area = (1/2) * AC * BC = (1/2) * 15 * 20 = (1/2) * 300 = 150Alternatively, using the hypotenuse and altitude:CH = √(AH * HB) = √(9 * 16) = √144 = 12Area = (1/2) * AB * CH = (1/2) * 25 * 12 = (1/2) * 300 = 150Both methods give the same result, so I think 150 is the correct area.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. I set AH = x, HB = 16, so AB = x + 16.2. Using the geometric mean for AC: AC² = AH * AB → 225 = x(x + 16)3. Solved quadratic equation: x² + 16x - 225 = 0, got x = 9.4. Then AB = 25, BC² = HB * AB = 16*25=400, so BC=20.5. Calculated area as (1/2)*15*20=150.Everything seems to check out. I think the area is 150.**Final Answer**boxed{C}

💡Okay, so I have this problem where there are n cinemas in a city, and k tourists have arrived. I need to figure out in how many different ways these tourists can distribute themselves among the cinemas. Hmm, let's see.First, I think about what it means for the tourists to distribute themselves. Each tourist can choose any of the n cinemas, right? So, for each tourist, there are n options. But since there are k tourists, I need to consider how their choices combine.Maybe I can think of it like this: if there's only one tourist, it's simple—there are n ways they can choose a cinema. Now, if there are two tourists, each can independently choose any of the n cinemas. So for the first tourist, there are n choices, and for the second tourist, also n choices. That means for two tourists, the total number of ways is n multiplied by n, which is n squared.Extending this idea, if there are three tourists, each with n choices, the total number of ways would be n times n times n, which is n cubed. So, it seems like for k tourists, the total number of ways would be n multiplied by itself k times, which is n to the power of k, or n^k.Wait, but is there a different way to think about this? Maybe using permutations or combinations? Let me consider that. If the tourists are distinguishable, meaning each one is unique, then the order in which they choose the cinemas matters. So, for each tourist, their choice is independent, and the total number of ways is indeed n^k.But what if the tourists are indistinct? Like, if we only care about how many tourists are in each cinema, not which specific tourists. That would be a different problem, right? In that case, it would be a problem of distributing k indistinct objects into n distinct boxes, which is a classic combinatorics problem. The formula for that is (n + k - 1) choose (k), or (n + k - 1) choose (n - 1). But in this problem, I think the tourists are considered distinct because they are different people, so their choices matter individually.Let me check with a small example to make sure. Suppose there are 2 cinemas and 2 tourists. According to my initial thought, the number of ways should be 2^2 = 4. Let's list them:1. Both tourists go to Cinema 1.2. Both tourists go to Cinema 2.3. Tourist 1 goes to Cinema 1, and Tourist 2 goes to Cinema 2.4. Tourist 1 goes to Cinema 2, and Tourist 2 goes to Cinema 1.Yes, that's 4 ways, which matches 2^2. If I tried using the combination formula for indistinct objects, it would be (2 + 2 - 1) choose 2 = 3 choose 2 = 3, which is less than 4. So, that confirms that the correct approach is to treat the tourists as distinct, leading to n^k ways.Another example: 3 cinemas and 1 tourist. The number of ways should be 3^1 = 3, which makes sense—each cinema is a possible choice for the single tourist.What about 1 cinema and 5 tourists? Then, there's only 1 way for all tourists to go to that single cinema, which is 1^5 = 1. That also makes sense.So, putting it all together, for each of the k tourists, there are n choices, and since each choice is independent, the total number of ways is n multiplied by itself k times, which is n^k.I think that's the correct approach. I don't see any flaws in this reasoning, and the examples I tested support this conclusion.