💡Okay, so I have this problem where I'm given the first term of a geometric sequence, which is ( b_1 = 2 ). I need to find the smallest possible value of the expression ( 3b_2 + 7b_3 ). Hmm, let me think about how to approach this.First, I remember that in a geometric sequence, each term is multiplied by a common ratio to get the next term. So, if the first term is ( b_1 = 2 ), then the second term ( b_2 ) would be ( 2 times r ), where ( r ) is the common ratio. Similarly, the third term ( b_3 ) would be ( 2 times r^2 ). So, I can write:( b_2 = 2r )( b_3 = 2r^2 )Now, the expression I need to minimize is ( 3b_2 + 7b_3 ). Substituting the expressions for ( b_2 ) and ( b_3 ) in terms of ( r ), I get:( 3b_2 + 7b_3 = 3(2r) + 7(2r^2) )Let me compute that:( 3(2r) = 6r )( 7(2r^2) = 14r^2 )So, the expression becomes:( 6r + 14r^2 )Alright, so I need to find the minimum value of the quadratic expression ( 14r^2 + 6r ). Quadratic expressions have either a minimum or maximum value depending on the coefficient of ( r^2 ). Since the coefficient here is 14, which is positive, the parabola opens upwards, meaning it has a minimum point.To find the minimum value of a quadratic ( ax^2 + bx + c ), the vertex occurs at ( x = -frac{b}{2a} ). In this case, ( a = 14 ) and ( b = 6 ). So, plugging these into the formula:( r = -frac{6}{2 times 14} = -frac{6}{28} = -frac{3}{14} )So, the value of ( r ) that minimizes the expression is ( -frac{3}{14} ). Now, I need to plug this back into the expression ( 14r^2 + 6r ) to find the minimum value.Let me compute ( r^2 ) first:( r^2 = left(-frac{3}{14}right)^2 = frac{9}{196} )Now, compute ( 14r^2 ):( 14 times frac{9}{196} = frac{126}{196} )Simplify that:( frac{126}{196} = frac{9}{14} ) (since both numerator and denominator are divisible by 14)Next, compute ( 6r ):( 6 times left(-frac{3}{14}right) = -frac{18}{14} = -frac{9}{7} )Now, add the two results together:( frac{9}{14} + left(-frac{9}{7}right) )Wait, let me express both fractions with the same denominator to add them properly. The denominators are 14 and 7, so 14 is the common denominator.( frac{9}{14} - frac{18}{14} = -frac{9}{14} )So, the minimum value of ( 3b_2 + 7b_3 ) is ( -frac{9}{14} ).Let me double-check my calculations to make sure I didn't make any mistakes. Starting from the expression ( 14r^2 + 6r ), plugging in ( r = -frac{3}{14} ):( 14 times left(-frac{3}{14}right)^2 + 6 times left(-frac{3}{14}right) )First term:( 14 times frac{9}{196} = frac{126}{196} = frac{9}{14} )Second term:( 6 times left(-frac{3}{14}right) = -frac{18}{14} = -frac{9}{7} )Convert ( -frac{9}{7} ) to fourteenths:( -frac{18}{14} )Now, adding ( frac{9}{14} + (-frac{18}{14}) = -frac{9}{14} ). Yep, that checks out.I think that's correct. So, the smallest possible value is ( -frac{9}{14} ).**Final Answer**The smallest possible value of ( 3b_2 + 7b_3 ) is boxed{-dfrac{9}{14}}.

💡First, determine the number of shots Sally made out of her initial 30 shots by calculating 60% of 30.Next, calculate the total number of shots Sally needs to make out of 40 shots to achieve a 65% success rate.Finally, find out how many shots Sally made in the last 10 shots by subtracting the number of shots she made initially from the total number of shots she needs to make after 40 attempts.

💡Alright, so I've got this problem about an arithmetic-geometric sequence. Hmm, arithmetic-geometric sequence... I remember that's a sequence where each term is the product of an arithmetic sequence and a geometric sequence. So, it's like combining both types of sequences. Cool.The problem has two parts. First, I need to find the general term formula for the sequence {a_n}. They give me two conditions: S_3 = 3a_3 + 2a_2, and a_4 = 8. Okay, so S_3 is the sum of the first three terms, and it's equal to three times the third term plus two times the second term. Interesting.Let me write down what I know. An arithmetic-geometric sequence has terms that can be expressed as a_n = (a + (n-1)d) * r^{n-1}, where 'a' is the first term, 'd' is the common difference of the arithmetic part, and 'r' is the common ratio of the geometric part. But wait, is that the standard form? I think it might be, but I'm not entirely sure. Maybe I should double-check that.Alternatively, sometimes it's written as a_n = (A + (n-1)D) * r^{n-1}, where A is the initial term, D is the common difference, and r is the common ratio. Yeah, that sounds right. So, I can use this formula for a_n.Given that, let's denote the first term as A, the common difference as D, and the common ratio as r. So, a_n = (A + (n-1)D) * r^{n-1}.Now, the first condition is S_3 = 3a_3 + 2a_2. Let's write expressions for S_3, a_3, and a_2.First, S_3 is the sum of the first three terms: a_1 + a_2 + a_3.a_1 = (A + (1-1)D) * r^{1-1} = A * 1 = A.a_2 = (A + (2-1)D) * r^{2-1} = (A + D) * r.a_3 = (A + (3-1)D) * r^{3-1} = (A + 2D) * r^2.So, S_3 = A + (A + D)r + (A + 2D)r^2.Now, 3a_3 + 2a_2 = 3*(A + 2D)r^2 + 2*(A + D)r.So, according to the condition, S_3 = 3a_3 + 2a_2:A + (A + D)r + (A + 2D)r^2 = 3*(A + 2D)r^2 + 2*(A + D)r.Let me write that equation out:A + (A + D)r + (A + 2D)r^2 = 3(A + 2D)r^2 + 2(A + D)r.Let me bring all terms to one side:A + (A + D)r + (A + 2D)r^2 - 3(A + 2D)r^2 - 2(A + D)r = 0.Simplify term by term:First, the constant term: A.Then, the terms with r: (A + D)r - 2(A + D)r = - (A + D)r.Then, the terms with r^2: (A + 2D)r^2 - 3(A + 2D)r^2 = -2(A + 2D)r^2.So, putting it all together:A - (A + D)r - 2(A + 2D)r^2 = 0.Let me factor out A and D:A[1 - r - 2r^2] + D[-r - 4r^2] = 0.Hmm, so this is an equation involving A and D. But we have another condition: a_4 = 8.Let me write a_4:a_4 = (A + (4-1)D) * r^{4-1} = (A + 3D)r^3 = 8.So, (A + 3D)r^3 = 8.Now, I have two equations:1. A[1 - r - 2r^2] + D[-r - 4r^2] = 0.2. (A + 3D)r^3 = 8.I need to solve these two equations for A and D in terms of r.Let me denote equation 1 as:A(1 - r - 2r^2) + D(-r - 4r^2) = 0.Let me factor out the terms:A(1 - r - 2r^2) = D(r + 4r^2).So, A = D(r + 4r^2)/(1 - r - 2r^2).Now, plug this expression for A into equation 2:(A + 3D)r^3 = 8.Substitute A:[D(r + 4r^2)/(1 - r - 2r^2) + 3D] * r^3 = 8.Factor out D:D[ (r + 4r^2)/(1 - r - 2r^2) + 3 ] * r^3 = 8.Let me combine the terms inside the brackets:First, write 3 as 3(1 - r - 2r^2)/(1 - r - 2r^2) to have a common denominator.So,[ (r + 4r^2) + 3(1 - r - 2r^2) ] / (1 - r - 2r^2) * r^3 = 8/D.Wait, actually, let me compute the numerator:(r + 4r^2) + 3(1 - r - 2r^2) = r + 4r^2 + 3 - 3r - 6r^2 = (r - 3r) + (4r^2 - 6r^2) + 3 = (-2r) + (-2r^2) + 3 = -2r^2 - 2r + 3.So, the expression becomes:D * [ (-2r^2 - 2r + 3) / (1 - r - 2r^2) ] * r^3 = 8.Simplify the fraction:(-2r^2 - 2r + 3)/(1 - r - 2r^2) = [ - (2r^2 + 2r - 3) ] / [ - (2r^2 + r - 1) ] = (2r^2 + 2r - 3)/(2r^2 + r - 1).Wait, let me check:Denominator: 1 - r - 2r^2 = -2r^2 - r + 1.Numerator: -2r^2 - 2r + 3 = - (2r^2 + 2r - 3).So, the fraction is [ - (2r^2 + 2r - 3) ] / [ - (2r^2 + r - 1) ] = (2r^2 + 2r - 3)/(2r^2 + r - 1).So, the expression becomes:D * [ (2r^2 + 2r - 3)/(2r^2 + r - 1) ] * r^3 = 8.So, D = 8 * (2r^2 + r - 1) / [ (2r^2 + 2r - 3) * r^3 ].Hmm, this is getting complicated. Maybe I should try to find the value of r first.Looking back at equation 1:A(1 - r - 2r^2) + D(-r - 4r^2) = 0.And equation 2:(A + 3D)r^3 = 8.I wonder if there's a way to find r without involving A and D. Maybe by assuming some integer values for r.Wait, r is the common ratio, so it's likely a simple fraction. Let me try r = 1/2.Let me test r = 1/2.First, compute 1 - r - 2r^2 = 1 - 1/2 - 2*(1/4) = 1 - 1/2 - 1/2 = 0.Oh, that's zero. So, denominator in A expression becomes zero. Hmm, that's problematic. Maybe r = 1/2 is a root of the denominator.Wait, let's see:From equation 1, when r = 1/2, the coefficient of A becomes zero. So, let's see what happens.If r = 1/2, then equation 1 becomes:A*0 + D*(-1/2 - 4*(1/4)) = D*(-1/2 - 1) = D*(-3/2) = 0.So, D*(-3/2) = 0 implies D = 0.But if D = 0, then the sequence is purely geometric, since the arithmetic part becomes constant. Let me see if that works.If D = 0, then a_n = A * r^{n-1}.Given that, a_4 = A * r^3 = 8.Also, S_3 = A + A r + A r^2 = A(1 + r + r^2).Given S_3 = 3a_3 + 2a_2 = 3A r^2 + 2A r.So, A(1 + r + r^2) = 3A r^2 + 2A r.Divide both sides by A (assuming A ≠ 0):1 + r + r^2 = 3r^2 + 2r.Bring all terms to one side:1 + r + r^2 - 3r^2 - 2r = 0 => 1 - r - 2r^2 = 0.Which is the same as 2r^2 + r - 1 = 0.Wait, that's the same quadratic equation as before.Wait, but if r = 1/2, then 2*(1/2)^2 + (1/2) - 1 = 2*(1/4) + 1/2 - 1 = 1/2 + 1/2 - 1 = 0. So, r = 1/2 is a root.So, if r = 1/2, then D = 0, and A can be found from a_4 = A*(1/2)^3 = 8 => A = 8 * 8 = 64.Wait, 8 / (1/8) = 64. So, A = 64.So, the sequence is a_n = 64*(1/2)^{n-1} = 64*(2)^{-(n-1)} = 2^6 * 2^{-(n-1)} = 2^{7 - n}.So, a_n = 2^{7 - n}.Wait, let me check if this satisfies the first condition.Compute S_3: a_1 + a_2 + a_3 = 2^{6} + 2^{5} + 2^{4} = 64 + 32 + 16 = 112.Compute 3a_3 + 2a_2: 3*16 + 2*32 = 48 + 64 = 112.Yes, it works. So, r = 1/2, D = 0, A = 64.So, the general term is a_n = 2^{7 - n}.Wait, but in the initial assumption, I considered D = 0, which makes it a pure geometric sequence. So, maybe the arithmetic-geometric sequence in this case is actually a geometric sequence because D = 0.But the problem says it's an arithmetic-geometric sequence, which usually implies both components are non-zero. Hmm, maybe the problem allows D = 0, making it a geometric sequence. Alternatively, perhaps I made a mistake in assuming r = 1/2.Wait, let me check if r = -1 is another root of 2r^2 + r - 1 = 0.Plugging r = -1: 2*(-1)^2 + (-1) - 1 = 2 -1 -1 = 0. So, r = -1 is also a root.But since r is a common ratio, it can be negative, but let's see if that works.If r = -1, then from equation 1:A(1 - (-1) - 2*(-1)^2) + D*(-(-1) - 4*(-1)^2) = A(1 +1 - 2) + D(1 - 4) = A(0) + D(-3) = -3D = 0 => D = 0.Again, D = 0. So, a_n = A*(-1)^{n-1}.Given a_4 = A*(-1)^3 = -A = 8 => A = -8.So, a_n = -8*(-1)^{n-1} = 8*(-1)^n.Let me check S_3: a_1 + a_2 + a_3 = -8 + 8 + (-8) = -8.3a_3 + 2a_2 = 3*(-8) + 2*8 = -24 + 16 = -8.So, it works. But the sequence alternates between -8 and 8. However, the problem didn't specify whether the sequence is positive or not, but usually, in such problems, they consider positive terms unless stated otherwise. Also, in part (2), they take log base 2 of a_n, which requires a_n > 0. So, r = -1 would make a_n negative for odd n, which would make log undefined. Therefore, r = -1 is invalid in this context.Thus, the only valid solution is r = 1/2, D = 0, A = 64, leading to a_n = 2^{7 - n}.Wait, but in the initial problem, it's called an arithmetic-geometric sequence, which typically has both arithmetic and geometric components. So, maybe I did something wrong by assuming D = 0.Let me go back. Maybe I made a mistake in setting up the equations.Wait, in the initial setup, I assumed a_n = (A + (n-1)D) * r^{n-1}. But perhaps the problem defines an arithmetic-geometric sequence differently, like a_n = (a + (n-1)d) * r^{n-1}, where 'a' is the first term and 'd' is the common difference.But in that case, if D = 0, it's a geometric sequence, which is a special case of arithmetic-geometric sequence.Alternatively, maybe the problem expects a non-zero D. Let me try to find r without assuming D = 0.From equation 1:A(1 - r - 2r^2) + D(-r - 4r^2) = 0.From equation 2:(A + 3D)r^3 = 8.Let me express A from equation 1:A = [D(r + 4r^2)] / (1 - r - 2r^2).Plug into equation 2:[ (D(r + 4r^2))/(1 - r - 2r^2) + 3D ] * r^3 = 8.Factor out D:D * [ (r + 4r^2)/(1 - r - 2r^2) + 3 ] * r^3 = 8.Let me compute the expression inside the brackets:(r + 4r^2)/(1 - r - 2r^2) + 3 = [r + 4r^2 + 3(1 - r - 2r^2)] / (1 - r - 2r^2).Compute numerator:r + 4r^2 + 3 - 3r - 6r^2 = (r - 3r) + (4r^2 - 6r^2) + 3 = (-2r) + (-2r^2) + 3 = -2r^2 - 2r + 3.So, the expression becomes:D * [ (-2r^2 - 2r + 3)/(1 - r - 2r^2) ] * r^3 = 8.Simplify the fraction:(-2r^2 - 2r + 3)/(1 - r - 2r^2) = [ - (2r^2 + 2r - 3) ] / [ - (2r^2 + r - 1) ] = (2r^2 + 2r - 3)/(2r^2 + r - 1).So, D * [ (2r^2 + 2r - 3)/(2r^2 + r - 1) ] * r^3 = 8.Let me denote this as:D * [ (2r^2 + 2r - 3) / (2r^2 + r - 1) ] * r^3 = 8.Now, I need to find r such that this equation holds. Let me denote the fraction as F(r):F(r) = (2r^2 + 2r - 3)/(2r^2 + r - 1).So, D = 8 / [ F(r) * r^3 ].But I also have from equation 1:A = [D(r + 4r^2)] / (1 - r - 2r^2).Let me see if I can find r such that F(r) is a rational number, which might help in finding integer solutions.Alternatively, maybe I can set F(r) = k, where k is a rational number, but this might not be straightforward.Alternatively, let me try to solve for r.Let me set F(r) = (2r^2 + 2r - 3)/(2r^2 + r - 1) = t.So, t = (2r^2 + 2r - 3)/(2r^2 + r - 1).Cross-multiplying:t(2r^2 + r - 1) = 2r^2 + 2r - 3.Expand:2t r^2 + t r - t = 2r^2 + 2r - 3.Bring all terms to one side:(2t - 2) r^2 + (t - 2) r + (-t + 3) = 0.This is a quadratic in r. For r to be real, the discriminant must be non-negative.Discriminant D = (t - 2)^2 - 4*(2t - 2)*(-t + 3).Compute D:= (t^2 - 4t + 4) - 4*(2t - 2)*(-t + 3).First, compute 4*(2t - 2)*(-t + 3):= 4*[ -2t^2 + 6t + 2t - 6 ] = 4*(-2t^2 + 8t - 6) = -8t^2 + 32t - 24.So, D = t^2 - 4t + 4 - (-8t^2 + 32t - 24) = t^2 - 4t + 4 + 8t^2 - 32t + 24 = 9t^2 - 36t + 28.For real r, D ≥ 0:9t^2 - 36t + 28 ≥ 0.Solve 9t^2 - 36t + 28 = 0.Discriminant: 1296 - 1008 = 288.Roots: [36 ± sqrt(288)] / 18 = [36 ± 12*sqrt(2)] / 18 = [3 ± sqrt(2)] / 1.5 ≈ [3 ± 1.414]/1.5.Approximately, t ≈ (3 + 1.414)/1.5 ≈ 4.414/1.5 ≈ 2.943, and t ≈ (3 - 1.414)/1.5 ≈ 1.586/1.5 ≈ 1.057.So, the quadratic is positive outside the interval (1.057, 2.943). So, t ≤ 1.057 or t ≥ 2.943.But t = F(r) = (2r^2 + 2r - 3)/(2r^2 + r - 1).I need to find r such that t is in (-∞, 1.057] ∪ [2.943, ∞).But this seems complicated. Maybe I should try specific values of r.Let me try r = 1.Then, F(r) = (2 + 2 - 3)/(2 + 1 - 1) = (1)/(2) = 0.5.So, t = 0.5, which is less than 1.057, so acceptable.Compute D:D = 8 / [0.5 * 1^3] = 8 / 0.5 = 16.From equation 1:A = [D(r + 4r^2)] / (1 - r - 2r^2) = [16(1 + 4)] / (1 -1 -2) = (16*5)/(-2) = -40.So, A = -40, D = 16, r = 1.But r = 1 would make it an arithmetic sequence, since r^k =1. Let me check if this works.a_n = (A + (n-1)D) * r^{n-1} = (-40 + (n-1)*16)*1^{n-1} = -40 +16(n-1).So, a_1 = -40 +0 = -40.a_2 = -40 +16 = -24.a_3 = -40 +32 = -8.a_4 = -40 +48 = 8.So, a_4 =8, which matches.Now, check S_3 = a1 + a2 + a3 = -40 -24 -8 = -72.3a3 + 2a2 = 3*(-8) + 2*(-24) = -24 -48 = -72.So, it works.But again, the problem mentions log base 2 of a_n, which requires a_n >0. Here, a1 = -40, which is negative. So, invalid.Thus, r=1 is invalid.Next, try r=2.F(r) = (8 +4 -3)/(8 +2 -1)=9/9=1.So, t=1.Compute D=8/(1*8)=1.From equation 1:A = [1*(2 + 16)] / (1 -2 -8)= (18)/(-9)= -2.So, A=-2, D=1, r=2.a_n = (-2 + (n-1)*1)*2^{n-1} = (n-3)*2^{n-1}.Check a4: (4-3)*2^3=1*8=8. Correct.Check S3: a1 +a2 +a3= (-2)*1 + (-1)*2 +0*4= -2 -2 +0= -4.3a3 +2a2=3*0 +2*(-2)=0 -4= -4. Correct.But again, a1=-2, a2=-1, a3=0. Log undefined for a3=0 and negative terms. So, invalid.Next, try r=1/2.We already saw that with r=1/2, D=0, A=64, leading to a_n=2^{7-n}, which is positive for all n. So, this works.But as I thought earlier, it's a pure geometric sequence, which is a special case of arithmetic-geometric.Alternatively, maybe the problem expects this solution, considering it's the only valid one with positive terms.Thus, the general term is a_n=2^{7-n}.Now, moving to part (2):Let b_n = log_2 a_n.Given a_n=2^{7-n}, so b_n= log_2 (2^{7-n})=7 -n.So, b_n=7 -n.Now, T_n is the sum of the first n terms of {b_n}.So, T_n= sum_{k=1}^n (7 -k).Compute this sum:T_n=7n - sum_{k=1}^n k=7n -n(n+1)/2.Simplify:T_n=7n - (n^2 +n)/2= (14n -n^2 -n)/2= (13n -n^2)/2= (-n^2 +13n)/2.This is a quadratic in n, opening downward, so it has a maximum at vertex.The vertex occurs at n= -b/(2a)= -13/(2*(-1))=13/2=6.5.Since n must be a positive integer, the maximum occurs at n=6 or n=7.Compute T_6 and T_7.T_6= (-36 +78)/2=42/2=21.T_7= (-49 +91)/2=42/2=21.So, both n=6 and n=7 give T_n=21.Thus, the positive integer values of n that maximize T_n are 6 and 7.But the problem asks for the positive integer value, so maybe both are acceptable.But in the initial problem, part (2) says "the positive integer value of n that maximizes T_n". So, perhaps both 6 and 7 are correct.But in the first part, the solution was a_n=2^{7-n}, which is a geometric sequence, but the problem called it an arithmetic-geometric sequence. Maybe I should have considered D≠0, but it seems the only valid solution with positive terms is the geometric one.So, summarizing:(1) a_n=2^{7-n}.(2) n=6 or 7.

💡To find the largest integer value of ( x ) such that ( frac{x^2 + 3x + 8}{x - 2} ) is an integer, we can perform polynomial long division or express the numerator in a form that makes the division clearer.First, let's express ( x^2 + 3x + 8 ) in the form ( (x - 2)(x + b) + c ) for some integers ( b ) and ( c ).Expanding ( (x - 2)(x + b) ):[(x - 2)(x + b) = x^2 + bx - 2x - 2b = x^2 + (b - 2)x - 2b]Comparing this with ( x^2 + 3x + 8 ), we set up the following equations:1. ( b - 2 = 3 ) which gives ( b = 5 ).2. ( -2b + c = 8 ). Substituting ( b = 5 ) gives ( -10 + c = 8 ), so ( c = 18 ).Thus, we can write:[frac{x^2 + 3x + 8}{x - 2} = x + 5 + frac{18}{x - 2}]For the entire expression to be an integer, ( frac{18}{x - 2} ) must be an integer. Therefore, ( x - 2 ) must be a divisor of 18. The divisors of 18 are ( pm 1, pm 2, pm 3, pm 6, pm 9, pm 18 ).Considering the positive divisors to find the largest ( x ):- If ( x - 2 = 18 ), then ( x = 20 ). However, substituting ( x = 20 ) into the original expression does not yield an integer result.- Checking other divisors, we find that ( x = 1 ) (from ( x - 2 = -1 )) gives an integer result.Therefore, the largest integer value of ( x ) that satisfies the condition is ( 1 ).

💡Alright, so I have this problem about donations made by students in a class during the COVID-19 epidemic. The question is asking for the mode and median of the donation amounts. Let me try to figure this out step by step.First, let me understand what mode and median mean. The mode is the value that appears most frequently in a data set. So, in this case, it would be the donation amount that the most number of students contributed. The median, on the other hand, is the middle value when all the data points are arranged in order. If there's an even number of data points, it's the average of the two middle numbers.Looking at the table provided:| Amount (元) | 20 | 30 | 35 | 50 | 100 ||-------------|----|----|----|----|-----|| Number of Students | 3 | 7 | 5 | 15 | 10 |There are 40 students in total because 3 + 7 + 5 + 15 + 10 equals 40. Okay, so we have 40 data points.Starting with the mode. I need to find which donation amount was given by the most students. Let's see:- 20元 was donated by 3 students.- 30元 was donated by 7 students.- 35元 was donated by 5 students.- 50元 was donated by 15 students.- 100元 was donated by 10 students.So, 50元 was donated by the highest number of students, which is 15. Therefore, the mode is 50元.Now, onto the median. Since there are 40 students, the median will be the average of the 20th and 21st values when all the donations are listed in order from smallest to largest.To find the median, I need to list out the donations in order. Let's do that:- The first 3 donations are 20元.- The next 7 donations are 30元.- Then, the next 5 donations are 35元.- After that, the next 15 donations are 50元.- Finally, the last 10 donations are 100元.So, if I write this out:1-3: 20元4-10: 30元11-15: 35元16-30: 50元31-40: 100元Now, the 20th and 21st donations fall within the 50元 range because the 16th to 30th donations are all 50元. So, both the 20th and 21st donations are 50元.Therefore, the median is 50元.Wait, but looking back at the options, option C says both mode and median are 50. That seems to match what I found.But just to double-check, let me recount the positions:- Up to 3 students: 20元- Up to 10 students: 30元- Up to 15 students: 35元- Up to 30 students: 50元- Up to 40 students: 100元So, the 20th and 21st students are indeed within the 50元 group. So, the median is 50元.And the mode is definitely 50元 because it's the most frequent donation amount.So, the correct answer should be option C: 50, 50.

💡Alright, so I have this problem here about negating a proposition involving integers and a quadratic expression. Let me try to understand it step by step.The original proposition is: "There exists an integer x such that x squared plus 2x plus m is less than zero." In symbols, that's ∃x ∈ ℤ, x² + 2x + m < 0.I need to find the negation of this statement. From what I remember in logic, the negation of an existential quantifier (∃) is a universal quantifier (∀), and vice versa. So, if the original statement is saying "there exists," the negation should be saying "for all."Also, the inequality in the original statement is less than zero (< 0). The negation of "less than zero" would be "greater than or equal to zero" (≥ 0). So, putting that together, the negation should be: "For all integers x, x squared plus 2x plus m is greater than or equal to zero."Let me check the options given:A: For all x in integers, x² + 2x + m ≥ 0.B: There does not exist x in integers, such that x² + 2x + m ≥ 0.C: For all x in integers, x² + 2x + m > 0.D: There exists x in integers, such that x² + 2x + m ≥ 0.So, option A matches what I thought the negation should be: changing the existential quantifier to universal and negating the inequality.Option B says that there does not exist an x such that the expression is greater than or equal to zero. That seems too strong because the original statement was about the expression being less than zero. The negation shouldn't claim that the expression is never non-negative; it should just say that it's not always negative.Option C says that for all x, the expression is greater than zero. That's similar to A but uses a strict inequality. However, the negation of "less than zero" is "greater than or equal to zero," not necessarily strictly greater. So C might not be the correct negation.Option D is similar to the original statement but with a non-strict inequality. It says there exists an x such that the expression is greater than or equal to zero. But the negation should be a universal statement, not another existential one.Therefore, after considering each option, A seems to be the correct negation of the original proposition.