💡Okay, so I have this inequality: ((a^{2}-1)x^{2}-(a-1)x-1 < 0). The question is asking for what value of (a) the solution set of this inequality is equal to all real numbers, which is (mathbb{R}). Hmm, that means no matter what (x) I plug in, the inequality should hold true. So, I need to find the values of (a) such that this quadratic inequality is always negative for every real (x).First, I remember that for a quadratic equation (Ax^2 + Bx + C), the graph is a parabola. If the parabola opens upwards (when (A > 0)), it will have a minimum point, and if it opens downwards (when (A < 0)), it will have a maximum point. For the quadratic to be always negative, it needs to open downwards and never touch or cross the x-axis. That means the quadratic must be negative definite.So, the conditions for a quadratic (Ax^2 + Bx + C) to be negative definite are:1. The leading coefficient (A < 0).2. The discriminant (B^2 - 4AC < 0).Let me apply these conditions to the given inequality.First, identify (A), (B), and (C) from the inequality:- (A = a^2 - 1)- (B = -(a - 1))- (C = -1)So, condition 1: (A < 0)That is, (a^2 - 1 < 0)Which simplifies to (a^2 < 1)Taking square roots, we get (-1 < a < 1).Condition 2: Discriminant (B^2 - 4AC < 0)Let me compute the discriminant:(B^2 = [-(a - 1)]^2 = (a - 1)^2)(4AC = 4(a^2 - 1)(-1) = -4(a^2 - 1))So, discriminant (= (a - 1)^2 - 4(a^2 - 1)(-1))Wait, hold on, actually, it's (B^2 - 4AC), so substituting:(= (a - 1)^2 - 4(a^2 - 1)(-1))Simplify this:First, expand ((a - 1)^2):(= a^2 - 2a + 1)Then, compute (4AC):(4(a^2 - 1)(-1) = -4(a^2 - 1) = -4a^2 + 4)So, the discriminant becomes:(a^2 - 2a + 1 - (-4a^2 + 4))Which is (a^2 - 2a + 1 + 4a^2 - 4)Combine like terms:(a^2 + 4a^2 = 5a^2)(-2a) remains as is.(1 - 4 = -3)So, discriminant (= 5a^2 - 2a - 3)We need this discriminant to be less than 0:(5a^2 - 2a - 3 < 0)Now, solve this quadratic inequality. Let's find the roots of the equation (5a^2 - 2a - 3 = 0).Using the quadratic formula:(a = frac{2 pm sqrt{(-2)^2 - 4*5*(-3)}}{2*5})Simplify inside the square root:(4 + 60 = 64)So, square root is 8.Thus, (a = frac{2 pm 8}{10})Compute both roots:1. (a = frac{2 + 8}{10} = frac{10}{10} = 1)2. (a = frac{2 - 8}{10} = frac{-6}{10} = -frac{3}{5})So, the roots are (a = 1) and (a = -frac{3}{5}).Since the coefficient of (a^2) in the discriminant quadratic is positive (5), the parabola opens upwards. Therefore, the inequality (5a^2 - 2a - 3 < 0) holds between the roots.Thus, the solution for the discriminant condition is:(-frac{3}{5} < a < 1)Now, combining this with the first condition ( -1 < a < 1), we need to find the intersection of these intervals.The interval from the discriminant is (-frac{3}{5} < a < 1), and the interval from the leading coefficient is (-1 < a < 1). The intersection is (-frac{3}{5} < a < 1).Wait, but hold on. When (a = 1), let's check the original inequality. If (a = 1), then (A = 1^2 - 1 = 0), so the quadratic becomes linear:((0)x^2 - (1 - 1)x - 1 = -1 < 0). So, this is always true. Therefore, (a = 1) is also a valid solution.But in our discriminant condition, (a = 1) makes the discriminant zero, right? Because when (a = 1), the discriminant is (5(1)^2 - 2(1) - 3 = 5 - 2 - 3 = 0). So, the quadratic would have a repeated root, meaning it touches the x-axis. But in our case, when (a = 1), the quadratic reduces to a linear equation, which is always negative. So, it's still valid.Therefore, we can include (a = 1) in our solution set.So, putting it all together, the values of (a) that satisfy the original inequality for all real (x) are:(-frac{3}{5} < a leq 1)Wait, but in the discriminant condition, we had ( -frac{3}{5} < a < 1), but when (a = 1), the inequality still holds because it reduces to a constant negative value. So, we can include (a = 1).But what about (a = -frac{3}{5})? Let me check. If (a = -frac{3}{5}), then the discriminant is zero, so the quadratic would touch the x-axis, meaning the inequality ((a^{2}-1)x^{2}-(a-1)x-1 < 0) would not hold for all (x), because at the root, it would be equal to zero, which doesn't satisfy the strict inequality. Therefore, (a = -frac{3}{5}) is excluded.So, the final range is (-frac{3}{5} < a leq 1).But wait, let me double-check when (a = 1). As I saw earlier, the inequality becomes (-1 < 0), which is always true, so (a = 1) is acceptable.What about when (a) is between (-frac{3}{5}) and 1? Let me pick a value, say (a = 0). Then the inequality becomes ((0 - 1)x^2 - (-1)x -1 < 0) which is (-x^2 + x -1 < 0). Let's see if this is always negative.Compute the discriminant: (B^2 - 4AC = (1)^2 - 4*(-1)*(-1) = 1 - 4 = -3 < 0). So, the quadratic is always negative, which is good.Another test: (a = frac{1}{2}). Then, (A = (frac{1}{2})^2 - 1 = frac{1}{4} - 1 = -frac{3}{4} < 0). The discriminant is (5*(frac{1}{2})^2 - 2*(frac{1}{2}) - 3 = 5*frac{1}{4} - 1 - 3 = frac{5}{4} - 4 = -frac{11}{4} < 0). So, the quadratic is always negative.What about (a = -frac{1}{2})? Then, (A = (-frac{1}{2})^2 - 1 = frac{1}{4} - 1 = -frac{3}{4} < 0). Discriminant: (5*(-frac{1}{2})^2 - 2*(-frac{1}{2}) - 3 = 5*frac{1}{4} + 1 - 3 = frac{5}{4} - 2 = -frac{3}{4} < 0). So, it's always negative.But wait, (a = -frac{3}{5}) is the boundary. Let me test (a = -frac{3}{5}). Then, (A = (-frac{3}{5})^2 - 1 = frac{9}{25} - 1 = -frac{16}{25} < 0). The discriminant is (5*(-frac{3}{5})^2 - 2*(-frac{3}{5}) - 3 = 5*frac{9}{25} + frac{6}{5} - 3 = frac{9}{5} + frac{6}{5} - 3 = frac{15}{5} - 3 = 3 - 3 = 0). So, discriminant is zero, which means the quadratic touches the x-axis. Therefore, the inequality is not strictly less than zero for all (x); it's equal to zero at one point. So, (a = -frac{3}{5}) is excluded.Similarly, if I take (a) just above (-frac{3}{5}), say (a = -frac{2}{5}), then discriminant is (5*(-frac{2}{5})^2 - 2*(-frac{2}{5}) - 3 = 5*frac{4}{25} + frac{4}{5} - 3 = frac{4}{5} + frac{4}{5} - 3 = frac{8}{5} - 3 = -frac{7}{5} < 0), so it's okay.What about (a) just below (-frac{3}{5}), say (a = -1)? Wait, (a = -1) is outside our interval. Let me check (a = -1). Then, (A = (-1)^2 - 1 = 0), so the inequality becomes linear: ((0)x^2 - (-2)x -1 = 2x -1 < 0). This is not true for all (x); for example, when (x = 1), it's (2 - 1 = 1 < 0) is false. So, (a = -1) is invalid, which is consistent with our solution.Another test: (a = -frac{4}{5}), which is less than (-frac{3}{5}). Then, discriminant is (5*(-frac{4}{5})^2 - 2*(-frac{4}{5}) - 3 = 5*frac{16}{25} + frac{8}{5} - 3 = frac{16}{5} + frac{8}{5} - 3 = frac{24}{5} - 3 = frac{24}{5} - frac{15}{5} = frac{9}{5} > 0). So, the discriminant is positive, meaning the quadratic crosses the x-axis, so the inequality is not always negative. Therefore, (a = -frac{4}{5}) is invalid, which is consistent.So, putting it all together, the values of (a) must satisfy (-frac{3}{5} < a leq 1). Therefore, the solution set is (left(-frac{3}{5}, 1right]).**Final Answer**The value of (a) is boxed{left( -dfrac{3}{5}, 1 right]}.

💡Okay, so I need to find the smallest four-digit integer that is a multiple of 5 and leaves specific remainders when divided by 11, 7, and 9. Let me break this down step by step.First, let's note the conditions:1. The number is a four-digit integer, so it must be between 1000 and 9999.2. It's a multiple of 5, which means it must end with either 0 or 5.3. When divided by 11, the remainder is 7.4. When divided by 7, the remainder is 4.5. When divided by 9, the remainder is 4.Hmm, okay. So, I need a number that satisfies all these conditions. Let me think about how to approach this.Since the number is a multiple of 5, it's either ending with 0 or 5. That might help narrow things down later. But first, maybe I can use the Chinese Remainder Theorem to solve the system of congruences given by the remainders when divided by 11, 7, and 9.Let me write down the congruences:1. ( x equiv 7 mod 11 )2. ( x equiv 4 mod 7 )3. ( x equiv 4 mod 9 )I notice that the remainders when divided by 7 and 9 are both 4. That might be helpful. Maybe I can combine these two congruences first.So, if ( x equiv 4 mod 7 ) and ( x equiv 4 mod 9 ), then ( x ) is congruent to 4 modulo the least common multiple (LCM) of 7 and 9. Since 7 and 9 are coprime, their LCM is 63. So, ( x equiv 4 mod 63 ).Now, I have two congruences:1. ( x equiv 7 mod 11 )2. ( x equiv 4 mod 63 )I need to find a number that satisfies both of these. Let me express ( x ) in terms of the second congruence: ( x = 63k + 4 ) for some integer ( k ).Now, substitute this into the first congruence:( 63k + 4 equiv 7 mod 11 )Let me simplify this. First, compute ( 63 mod 11 ). 11 times 5 is 55, so 63 - 55 is 8. Therefore, ( 63 equiv 8 mod 11 ). So, the equation becomes:( 8k + 4 equiv 7 mod 11 )Subtract 4 from both sides:( 8k equiv 3 mod 11 )Now, I need to solve for ( k ). This means I need the multiplicative inverse of 8 modulo 11. Let me find a number ( y ) such that ( 8y equiv 1 mod 11 ).Trying y=1: 8*1=8≡8≠1y=2: 16≡5≠1y=3:24≡2≠1y=4:32≡10≠1y=5:40≡7≠1y=6:48≡4≠1y=7:56≡1. Oh, y=7 works because 8*7=56≡1 mod11.So, the inverse of 8 modulo11 is 7. Therefore, multiply both sides of the equation ( 8k equiv 3 mod 11 ) by 7:( k equiv 3*7 mod 11 )( k equiv 21 mod 11 )21 divided by 11 is 1 with remainder 10, so ( k equiv 10 mod 11 ).This means ( k = 11m + 10 ) for some integer ( m ).Now, substitute back into ( x = 63k + 4 ):( x = 63*(11m + 10) + 4 )( x = 693m + 630 + 4 )( x = 693m + 634 )So, the general solution is ( x = 693m + 634 ). Now, I need to find the smallest four-digit number in this sequence.Let's find the smallest ( m ) such that ( x geq 1000 ).So, ( 693m + 634 geq 1000 )Subtract 634: ( 693m geq 366 )Divide by 693: ( m geq 366 / 693 approx 0.528 )Since ( m ) must be an integer, the smallest ( m ) is 1.So, plugging ( m = 1 ):( x = 693*1 + 634 = 693 + 634 = 1327 )Wait, 1327 is a four-digit number, but I need to check if it's a multiple of 5. 1327 ends with a 7, so it's not a multiple of 5. Hmm, that's a problem.So, I need to find the next value of ( m ) such that ( x ) is a multiple of 5. Let's see.The next value would be ( m = 2 ):( x = 693*2 + 634 = 1386 + 634 = 2020 )2020 ends with a 0, so it's a multiple of 5. Let me check if it satisfies all the original conditions.Check divisibility by 11: 2020 divided by 11. 11*183=2013, so 2020 - 2013=7. So, remainder 7. Good.Divided by 7: 2020 /7. 7*288=2016, so 2020-2016=4. Remainder 4. Good.Divided by 9: 2020 /9. 9*224=2016, so 2020-2016=4. Remainder 4. Good.And it's a multiple of 5. Perfect.But wait, is 2020 the smallest four-digit number that satisfies all these conditions? Let me check if there's a smaller four-digit number in the sequence.We saw that for ( m=1 ), x=1327, which is not a multiple of 5. The next one is 2020. Is there any number between 1000 and 2020 that fits?Wait, let's see. The general solution is x=693m +634. So, starting from m=0, x=634 (three-digit), m=1, x=1327, m=2, x=2020.So, the next number after 634 is 1327, which is four-digit but not a multiple of 5. Then, 2020 is the next one, which is a multiple of 5. So, 2020 is indeed the smallest four-digit number that satisfies all the conditions.Therefore, the answer is 2020.

💡To find the value of ( A + B + C + D ) for the equation ((x + 3)(4x^2 - 2x + 7) = Ax^3 + Bx^2 + Cx + D), I will start by expanding the left-hand side.First, I'll distribute each term in the binomial ( (x + 3) ) across the trinomial ( (4x^2 - 2x + 7) ):1. Multiply ( x ) by each term in the trinomial: - ( x times 4x^2 = 4x^3 ) - ( x times (-2x) = -2x^2 ) - ( x times 7 = 7x )2. Multiply ( 3 ) by each term in the trinomial: - ( 3 times 4x^2 = 12x^2 ) - ( 3 times (-2x) = -6x ) - ( 3 times 7 = 21 )Next, I'll combine all these terms:[4x^3 - 2x^2 + 7x + 12x^2 - 6x + 21]Now, I'll combine like terms:- Combine the ( x^2 ) terms: ( -2x^2 + 12x^2 = 10x^2 )- Combine the ( x ) terms: ( 7x - 6x = x )This simplifies the expression to:[4x^3 + 10x^2 + x + 21]From this, I can identify the coefficients:- ( A = 4 )- ( B = 10 )- ( C = 1 )- ( D = 21 )Finally, I'll calculate the sum ( A + B + C + D ):[4 + 10 + 1 + 21 = 36]

💡Okay, so I have this problem about an arithmetic sequence. Let me see... It says that in the sequence {a_n}, a_6 plus a_8 equals 16, and a_4 is 1. I need to find the value of a_10. The options are A:15, B:30, C:31, D:64.Hmm, arithmetic sequences. I remember they have a common difference between consecutive terms. So, the general formula for the nth term of an arithmetic sequence is a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference.Alright, so let me write down what I know:1. a_6 + a_8 = 162. a_4 = 1I need to find a_10.Let me express each of these terms using the general formula.First, a_6 = a_1 + (6-1)d = a_1 + 5dSimilarly, a_8 = a_1 + (8-1)d = a_1 + 7dSo, a_6 + a_8 = (a_1 + 5d) + (a_1 + 7d) = 2a_1 + 12d = 16That's equation one: 2a_1 + 12d = 16Next, a_4 = a_1 + (4-1)d = a_1 + 3d = 1That's equation two: a_1 + 3d = 1So now I have two equations:1. 2a_1 + 12d = 162. a_1 + 3d = 1I can solve these two equations to find a_1 and d.Let me solve equation two for a_1:a_1 = 1 - 3dNow, substitute this into equation one:2(1 - 3d) + 12d = 16Let me compute that:2*1 = 22*(-3d) = -6dSo, 2 - 6d + 12d = 16Combine like terms:2 + 6d = 16Subtract 2 from both sides:6d = 14Divide both sides by 6:d = 14/6 = 7/3 ≈ 2.333...Hmm, that's a fraction. Is that okay? I guess so, arithmetic sequences can have fractional differences.Now, plug d back into equation two to find a_1:a_1 = 1 - 3*(7/3) = 1 - 7 = -6So, a_1 is -6 and d is 7/3.Now, I need to find a_10.Using the general formula:a_10 = a_1 + (10 - 1)d = a_1 + 9dSubstitute the values:a_10 = -6 + 9*(7/3)Compute 9*(7/3): 9 divided by 3 is 3, so 3*7 = 21So, a_10 = -6 + 21 = 15Wait, 15 is option A. So, the answer should be A:15.But let me double-check my calculations to make sure I didn't make a mistake.First, a_6 + a_8 = 16.Compute a_6: a_1 + 5d = -6 + 5*(7/3) = -6 + 35/3 = (-18/3 + 35/3) = 17/3 ≈5.666...Compute a_8: a_1 + 7d = -6 + 7*(7/3) = -6 + 49/3 = (-18/3 + 49/3) = 31/3 ≈10.333...Add them together: 17/3 + 31/3 = 48/3 = 16. That checks out.Also, a_4 = a_1 + 3d = -6 + 3*(7/3) = -6 + 7 = 1. That also checks out.So, a_10 is indeed 15. So, the correct answer is A:15.Wait, but I just thought, is there another way to approach this problem without solving for a_1 and d? Maybe using properties of arithmetic sequences.I remember that in an arithmetic sequence, the sum of terms equidistant from the ends is constant. But here, a_6 and a_8 aren't equidistant from the ends unless we know the total number of terms, which we don't.Alternatively, I recall that a_m + a_n = a_p + a_q if m + n = p + q. So, in this case, 6 + 8 = 14, and 4 + 10 = 14. So, a_6 + a_8 = a_4 + a_10.So, 16 = 1 + a_10, which gives a_10 = 15. That's a quicker way without having to solve for a_1 and d.Yeah, that's a smarter approach. I should remember that property for future problems. It can save time.So, either way, whether by solving the system of equations or using the property of arithmetic sequences, I get a_10 = 15. So, the answer is A:15.**Final Answer**boxed{A}

💡Alright, so I've got this math problem here, and I need to figure out how many rats Bill bought. Let me read it again to make sure I understand it correctly."Bill thought he bought 70 chihuahuas, but some of them turned out to be rats. If the number of rats was 6 times the number of chihuahuas, how many rats did he buy?"Okay, so Bill intended to buy 70 chihuahuas, but some of those were actually rats. The number of rats is six times the number of chihuahuas. I need to find out how many rats he bought.Let me break this down. There are two types of animals here: chihuahuas and rats. The total number of animals Bill thought he bought is 70. However, some of those are rats, not chihuahuas. So, the actual number of chihuahuas is less than 70, and the rest are rats.The problem states that the number of rats is six times the number of chihuahuas. So, if I let the number of chihuahuas be a certain value, the number of rats would be six times that value.Let me assign some variables to make this clearer. Let's say:- Let ( c ) represent the number of chihuahuas.- Let ( r ) represent the number of rats.According to the problem, the total number of animals is 70. So, the sum of chihuahuas and rats should be 70. That gives me the first equation:[ c + r = 70 ]The second piece of information is that the number of rats is six times the number of chihuahuas. So, the number of rats ( r ) is equal to six times the number of chihuahuas ( c ). That gives me the second equation:[ r = 6c ]Now, I have a system of two equations:1. ( c + r = 70 )2. ( r = 6c )Since I have two equations and two variables, I can solve this system to find the values of ( c ) and ( r ).Let me substitute the second equation into the first one. That means wherever I see ( r ) in the first equation, I'll replace it with ( 6c ).So, substituting ( r = 6c ) into ( c + r = 70 ):[ c + 6c = 70 ]Now, combining like terms:[ 7c = 70 ]To find the value of ( c ), I'll divide both sides of the equation by 7:[ c = frac{70}{7} ][ c = 10 ]So, Bill bought 10 chihuahuas. Now, to find out how many rats he bought, I'll use the second equation ( r = 6c ).Substituting ( c = 10 ) into ( r = 6c ):[ r = 6 times 10 ][ r = 60 ]Therefore, Bill bought 60 rats.Let me double-check my work to make sure I didn't make any mistakes. I set up the equations correctly based on the problem statement. I substituted the second equation into the first one properly, combined like terms, and solved for ( c ). Then, I used that value to find ( r ). The calculations seem straightforward, and I don't see any errors in the steps I took.Just to be thorough, I'll verify the solution by plugging the values back into the original equations.First equation: ( c + r = 70 )[ 10 + 60 = 70 ][ 70 = 70 ]That checks out.Second equation: ( r = 6c )[ 60 = 6 times 10 ][ 60 = 60 ]That also checks out.Everything seems consistent. So, I'm confident that the number of rats Bill bought is 60.

💡Okay, so I have this problem about an arithmetic sequence. Let me try to understand it step by step. The problem says that the sum of the first n terms of an arithmetic sequence {a_n} is S_n. It also tells me that a_1 is greater than 0, which means the first term is positive. Additionally, it's given that S_{36} equals S_{10}. I need to find the largest term in the sequence {S_n}, and the options are S_{10}, S_{23}, S_{24}, or S_{36}.Hmm, let me recall some properties of arithmetic sequences and their sums. The sum of the first n terms of an arithmetic sequence can be calculated using the formula:S_n = (n/2) * [2a_1 + (n - 1)d]where a_1 is the first term and d is the common difference.Given that S_{36} = S_{10}, I can set up the equation:(36/2) * [2a_1 + 35d] = (10/2) * [2a_1 + 9d]Simplifying both sides:18 * [2a_1 + 35d] = 5 * [2a_1 + 9d]Let me compute each side:Left side: 18*(2a1 + 35d) = 36a1 + 630dRight side: 5*(2a1 + 9d) = 10a1 + 45dSo, setting them equal:36a1 + 630d = 10a1 + 45dSubtracting 10a1 and 45d from both sides:26a1 + 585d = 0Hmm, that's 26a1 = -585dSo, a1 = (-585/26)dSimplify that fraction: 585 divided by 26. Let me see, 26*22 = 572, so 585 - 572 = 13, so 585/26 = 22.5. Wait, 26*22.5 = 585? Let me check:26*22 = 572, 26*0.5 = 13, so 572 +13=585. Yes, so 585/26=22.5.Therefore, a1 = -22.5dBut we know that a1 > 0, so -22.5d > 0, which implies that d < 0. So the common difference is negative. That means the arithmetic sequence is decreasing. Each term is smaller than the previous one.Now, since the sequence is decreasing, the terms will eventually become negative. So, the sum S_n will increase initially as we add positive terms, but after a certain point, adding negative terms will cause the sum to decrease. Therefore, the maximum sum S_n occurs at the point where the terms are still positive, just before they become negative.So, to find the maximum S_n, I need to find the value of n where a_n is still positive, but a_{n+1} is negative. That is, the last positive term in the sequence.Given that a1 = -22.5d, and d is negative, let me denote d as -k where k is positive. So, d = -k, and a1 = 22.5k.So, the nth term of the arithmetic sequence is:a_n = a1 + (n - 1)d = 22.5k + (n - 1)(-k) = 22.5k - (n - 1)k = (22.5 - (n - 1))kWe want to find the largest n such that a_n > 0.So, (22.5 - (n - 1))k > 0Since k is positive, we can divide both sides by k:22.5 - (n - 1) > 0Which simplifies to:22.5 > n - 1So,n - 1 < 22.5Therefore,n < 23.5Since n must be an integer, the largest n where a_n is positive is n = 23.Therefore, the 23rd term is the last positive term, and the 24th term is negative.Hence, the sum S_n will be maximum at n = 23 because adding the 24th term, which is negative, will start decreasing the sum.Wait, but let me verify this because sometimes the maximum can be at n = 24 if the 24th term is still positive or if the average is considered.Alternatively, another approach is to consider the sum S_n as a quadratic function in terms of n. Since S_n is a quadratic function, its graph is a parabola. Given that the coefficient of n^2 is (d/2), which is negative because d is negative, the parabola opens downward, meaning the vertex is the maximum point.The vertex of a parabola given by S_n = (d/2)n^2 + (a1 - d/2)n occurs at n = -b/(2a), where a is the coefficient of n^2 and b is the coefficient of n.In this case, a = d/2 and b = a1 - d/2.So, n = -(a1 - d/2)/(2*(d/2)) = -(a1 - d/2)/dPlugging in a1 = -22.5d:n = -(-22.5d - d/2)/d = (22.5d + 0.5d)/d = 23d/d = 23So, the maximum occurs at n = 23.Therefore, S_{23} is the maximum sum.Wait, but earlier I thought it might be 24, but using the vertex formula, it's 23. So, I think 23 is correct.Let me check another way. Since S_{36} = S_{10}, the sum from term 11 to 36 is zero. That is, a_{11} + a_{12} + ... + a_{36} = 0.The number of terms from 11 to 36 is 26 terms. So, the average of these 26 terms is zero. In an arithmetic sequence, the average of consecutive terms is equal to the average of the first and last term in that range.So, (a_{11} + a_{36}) / 2 = 0 => a_{11} + a_{36} = 0.But in an arithmetic sequence, a_{36} = a_{11} + 25d.So, a_{11} + a_{11} + 25d = 0 => 2a_{11} + 25d = 0 => a_{11} = -25d/2.But since a1 = -22.5d, let's express a_{11}:a_{11} = a1 + 10d = -22.5d + 10d = -12.5dSo, -12.5d = -25d/2, which is consistent because -25d/2 = -12.5d.So, that checks out.Now, since a_{11} = -12.5d and d is negative, a_{11} is positive because d is negative, so -12.5d is positive.Similarly, a_{36} = a1 + 35d = -22.5d + 35d = 12.5d, which is negative because d is negative.So, the terms from 11 to 36 start positive and end negative, crossing zero somewhere in between.Therefore, the maximum sum occurs just before the terms become negative. Since a_{23} is the last positive term, as we found earlier, the sum up to 23 is the maximum.Alternatively, let me compute a_{23} and a_{24}:a_{23} = a1 + 22d = -22.5d + 22d = (-22.5 + 22)d = (-0.5)dSince d is negative, (-0.5)d is positive because negative times negative is positive.a_{24} = a1 + 23d = -22.5d + 23d = (0.5)dSince d is negative, 0.5d is negative.So, a_{23} is positive, a_{24} is negative. Therefore, the sum S_{23} is the last sum before adding negative terms, so it's the maximum.Therefore, the answer should be S_{23}, which is option B.Wait, but let me make sure I didn't make a mistake in calculating a_{23}.a1 = -22.5da_{23} = a1 + 22d = -22.5d + 22d = (-22.5 + 22)d = (-0.5)dYes, that's correct. Since d is negative, (-0.5)d is positive.Similarly, a_{24} = a1 + 23d = -22.5d + 23d = 0.5d, which is negative.So, yes, S_{23} is the maximum.Alternatively, let me think about the average of the terms. The sum S_n is equal to n times the average of the first and nth term.So, S_n = n*(a1 + a_n)/2Since a_n = a1 + (n-1)d, this becomes S_n = n*(2a1 + (n-1)d)/2, which is the same as the formula I used earlier.Given that, the maximum occurs when the average of the first and nth term is maximized.Since the sequence is decreasing, the average will increase until the nth term is still positive, then start decreasing when the nth term becomes negative.Therefore, the maximum average occurs when a_n is just positive, which is at n=23.Hence, S_{23} is the maximum.I think that's solid. So, the answer is B: S_{23}.