💡Okay, so I have this problem that says if a plus b plus c equals zero, then a cubed plus b cubed plus c cubed equals three times a times b times c. Hmm, that seems interesting. I remember learning some algebraic identities, but I'm not exactly sure how to approach this one. Let me think about it step by step.First, the given condition is a + b + c = 0. Maybe I can use this to express one of the variables in terms of the others. Let's say I solve for c: c = -a - b. That might be useful later on.Now, I need to find a relationship between a³ + b³ + c³ and abc. I recall that there's an identity for the sum of cubes: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca). Wait, that looks promising because if a + b + c is zero, then the left side of the equation might simplify.Let me write that identity down:a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)Since a + b + c = 0, the right side becomes zero. So, the equation simplifies to:a³ + b³ + c³ - 3abc = 0Which means:a³ + b³ + c³ = 3abcOh, that's exactly what I needed to prove! So, using the sum of cubes identity and the given condition that a + b + c = 0, I was able to show that a³ + b³ + c³ equals 3abc.Wait, but let me double-check to make sure I didn't make any mistakes. Maybe I can try plugging in some numbers to see if it works. Let's choose a = 1, b = -1, and c = 0. Then a + b + c = 1 - 1 + 0 = 0, which satisfies the condition.Now, let's compute a³ + b³ + c³: 1³ + (-1)³ + 0³ = 1 - 1 + 0 = 0.And 3abc: 3 * 1 * (-1) * 0 = 0.So, both sides are equal, which checks out.Let me try another set of numbers. How about a = 2, b = -1, and c = -1. Then a + b + c = 2 - 1 - 1 = 0.Compute a³ + b³ + c³: 8 + (-1) + (-1) = 6.And 3abc: 3 * 2 * (-1) * (-1) = 6.Again, both sides are equal. That seems to work.Maybe one more example to be sure. Let's take a = 3, b = -2, and c = -1. Then a + b + c = 3 - 2 - 1 = 0.Compute a³ + b³ + c³: 27 + (-8) + (-1) = 18.And 3abc: 3 * 3 * (-2) * (-1) = 18.Consistently, both sides are equal. This gives me more confidence that the proof is correct.I think I understand now. The key was recognizing the sum of cubes identity and realizing that since a + b + c = 0, the entire expression simplifies neatly. It's a clever way to relate the sum of cubes to the product of the variables. I should remember this identity for future problems.Another thing I noticed is that this identity is a special case where the sum of the variables is zero. It might not hold if a + b + c isn't zero, but in this specific scenario, it works perfectly. It's interesting how certain conditions can simplify complex expressions.I wonder if there are other similar identities or if this one is part of a broader set of formulas. Maybe if I explore more algebraic identities, I can find other useful relationships like this one. It's always helpful to have a toolkit of identities to apply when solving problems.In summary, by using the sum of cubes identity and substituting the given condition that a + b + c = 0, I was able to prove that a³ + b³ + c³ equals 3abc. The numerical examples I tried also supported this conclusion, which makes me feel confident about the proof.

💡Okay, so I have this problem about a sequence defined by two initial primes, p and q, where p is less than q. The rule for generating the next term is a bit tricky: if the sum of the current two terms is a power of two, then the next term is 2; otherwise, it's the smallest odd prime factor of that sum. I need to show that after some point, the sequence becomes periodic with the cycle 2, 7, 3, 5.First, I should probably start by understanding how the sequence behaves. Let me take some small primes for p and q and see what happens. Let's say p = 2 and q = 3. Then:a1 = 2, a2 = 3.Now, a1 + a2 = 5, which is not a power of two. The smallest odd prime factor of 5 is 5 itself, so a3 = 5.Next, a2 + a3 = 3 + 5 = 8, which is 2^3. So, a4 = 2.Now, a3 + a4 = 5 + 2 = 7, which is not a power of two. The smallest odd prime factor of 7 is 7, so a5 = 7.Then, a4 + a5 = 2 + 7 = 9, which is not a power of two. The smallest odd prime factor of 9 is 3, so a6 = 3.Next, a5 + a6 = 7 + 3 = 10, which is not a power of two. The smallest odd prime factor of 10 is 5, so a7 = 5.Then, a6 + a7 = 3 + 5 = 8, which is 2^3. So, a8 = 2.Hmm, I see a pattern here: 2, 3, 5, 2, 7, 3, 5, 2, 7, 3, 5, 2,... It seems like after a certain point, it cycles through 2, 7, 3, 5 repeatedly.Wait, let me check another starting point. Suppose p = 3 and q = 5.a1 = 3, a2 = 5.a1 + a2 = 8, which is 2^3, so a3 = 2.a2 + a3 = 5 + 2 = 7, which is not a power of two. The smallest odd prime factor of 7 is 7, so a4 = 7.a3 + a4 = 2 + 7 = 9, which is not a power of two. The smallest odd prime factor of 9 is 3, so a5 = 3.a4 + a5 = 7 + 3 = 10, which is not a power of two. The smallest odd prime factor of 10 is 5, so a6 = 5.a5 + a6 = 3 + 5 = 8, which is 2^3, so a7 = 2.Again, the cycle 2, 7, 3, 5 starts appearing.Maybe this cycle is inevitable once we hit 2. Let me try starting with p = 5 and q = 7.a1 = 5, a2 = 7.a1 + a2 = 12, which is not a power of two. The smallest odd prime factor of 12 is 3, so a3 = 3.a2 + a3 = 7 + 3 = 10, which is not a power of two. The smallest odd prime factor of 10 is 5, so a4 = 5.a3 + a4 = 3 + 5 = 8, which is 2^3, so a5 = 2.a4 + a5 = 5 + 2 = 7, which is not a power of two. The smallest odd prime factor of 7 is 7, so a6 = 7.a5 + a6 = 2 + 7 = 9, which is not a power of two. The smallest odd prime factor of 9 is 3, so a7 = 3.a6 + a7 = 7 + 3 = 10, which is not a power of two. The smallest odd prime factor of 10 is 5, so a8 = 5.a7 + a8 = 3 + 5 = 8, which is 2^3, so a9 = 2.Again, the cycle 2, 7, 3, 5 is repeating.It seems that once the sequence hits 2, it starts cycling through 2, 7, 3, 5. So maybe the key is to show that the sequence will eventually reach 2, and then it cycles.How can I show that the sequence will eventually reach 2? Well, if the sequence doesn't reach 2, then all terms are odd primes. But the sum of two odd primes is even, so it's either a power of two or has some odd prime factors.If the sum is a power of two, then the next term is 2. If not, the next term is the smallest odd prime factor of the sum. But if we keep getting odd primes, the sequence might decrease or something?Wait, let's think about it. Suppose we have two odd primes, p and q. Their sum is even, so it's either a power of two or has some odd prime factors. If it's a power of two, we get 2. If not, we get the smallest odd prime factor, which is at least 3.But if we keep getting terms that are odd primes, the sequence might not necessarily decrease. For example, starting with 3 and 5:3, 5, 2, 7, 3, 5, 2, 7, 3, 5, 2,...But if I start with larger primes, say 7 and 11:a1 = 7, a2 = 11.a1 + a2 = 18, which is not a power of two. The smallest odd prime factor of 18 is 3, so a3 = 3.a2 + a3 = 11 + 3 = 14, which is not a power of two. The smallest odd prime factor of 14 is 7, so a4 = 7.a3 + a4 = 3 + 7 = 10, which is not a power of two. The smallest odd prime factor of 10 is 5, so a5 = 5.a4 + a5 = 7 + 5 = 12, which is not a power of two. The smallest odd prime factor of 12 is 3, so a6 = 3.a5 + a6 = 5 + 3 = 8, which is 2^3, so a7 = 2.a6 + a7 = 3 + 2 = 5, which is not a power of two. The smallest odd prime factor of 5 is 5, so a8 = 5.a7 + a8 = 2 + 5 = 7, which is not a power of two. The smallest odd prime factor of 7 is 7, so a9 = 7.a8 + a9 = 5 + 7 = 12, which is not a power of two. The smallest odd prime factor of 12 is 3, so a10 = 3.a9 + a10 = 7 + 3 = 10, which is not a power of two. The smallest odd prime factor of 10 is 5, so a11 = 5.a10 + a11 = 3 + 5 = 8, which is 2^3, so a12 = 2.And then it cycles again: 2, 7, 3, 5, 2, 7, 3, 5,...So even with larger starting primes, the sequence eventually reaches 2 and then cycles.I think the key is that the sequence cannot continue indefinitely without hitting a power of two. Because if it doesn't, the terms would have to keep getting smaller or something, but primes can't get smaller indefinitely. Wait, but primes can be large. Hmm.Alternatively, maybe the sequence must eventually reach a sum that is a power of two. Because if you keep taking the smallest prime factors, you might eventually get to a point where the sum is a power of two.But how can I formalize that? Maybe by considering that the sequence of sums either eventually hits a power of two or cycles through some primes. But if it cycles through primes, the sums might not be powers of two, but the terms would have to repeat, leading to a cycle.Wait, but in the examples I tried, once it hits 2, it cycles through 2, 7, 3, 5. So maybe the cycle is inevitable once 2 is reached.So perhaps I can argue that the sequence must eventually reach 2, and once it does, it cycles through 2, 7, 3, 5.To show that the sequence must reach 2, I can use the fact that the sequence cannot have all terms greater than 2. Because if it did, then all terms are odd primes, and their sums are even numbers greater than 2, which are either powers of two or have odd prime factors.But if the sum is a power of two, then the next term is 2. If not, the next term is the smallest odd prime factor, which is at least 3.But if the sequence never reaches 2, then all terms are odd primes, and the sums are even numbers greater than 2, which are either powers of two or have odd prime factors.But if the sum is a power of two, we get 2, which contradicts the assumption that all terms are odd primes. So the sequence must reach 2 eventually.Once 2 is reached, let's see what happens. Suppose a_n = 2 and a_{n+1} = x, where x is a prime.Then, a_n + a_{n+1} = 2 + x. If 2 + x is a power of two, then a_{n+2} = 2. Otherwise, a_{n+2} is the smallest odd prime factor of 2 + x.But 2 + x is an odd number if x is odd, which it is since x is a prime greater than 2. So 2 + x is odd, so it can't be a power of two (since powers of two are even except for 1, which is not relevant here). Therefore, a_{n+2} is the smallest odd prime factor of 2 + x.But 2 + x is an odd number greater than 2, so it has at least one odd prime factor. The smallest one is at least 3.So, for example, if x = 7, then 2 + 7 = 9, whose smallest prime factor is 3. So a_{n+2} = 3.Then, a_{n+1} + a_{n+2} = 7 + 3 = 10, which is not a power of two. The smallest odd prime factor of 10 is 5, so a_{n+3} = 5.Then, a_{n+2} + a_{n+3} = 3 + 5 = 8, which is 2^3, so a_{n+4} = 2.And then the cycle repeats: 2, 7, 3, 5, 2, 7, 3, 5,...So once we hit 2, the sequence cycles through 2, 7, 3, 5.Therefore, the sequence must eventually reach 2, and once it does, it cycles through 2, 7, 3, 5 indefinitely.I think that's the gist of the proof. I need to formalize it, maybe using induction or some lemma about the sequence eventually reaching 2.

💡Okay, so I have this problem here where I need to prove that among a set of real numbers ( x_1, x_2, ldots, x_n ), there exist two numbers ( x_p ) and ( x_q ) such that the inequality ( |a - b| leq |a - x_p| leq |x_q - x_p| ) holds. Here, ( a ) and ( b ) are weighted averages of these ( x_i )s with positive integer weights ( a_i ) and ( b_i ) respectively.First, let me make sure I understand the definitions correctly. The value ( a ) is a weighted average where each ( x_i ) is multiplied by its corresponding weight ( a_i ), and then divided by the sum of all ( a_i )s. Similarly, ( b ) is another weighted average using weights ( b_i ). So both ( a ) and ( b ) are kind of central tendencies of the ( x_i )s, but with different weights.The goal is to show that the difference between these two averages ( |a - b| ) is less than or equal to the distance from ( a ) to some ( x_p ), which in turn is less than or equal to the maximum distance between any two ( x_i )s, specifically ( |x_q - x_p| ).Hmm, okay. So, I need to find two specific ( x_p ) and ( x_q ) such that these inequalities hold. Maybe I can approach this by considering the differences between ( a ) and each ( x_i ), and similarly between ( b ) and each ( x_i ).Let me start by writing out the expressions for ( a ) and ( b ) more formally:[a = frac{sum_{i=1}^{n} a_i x_i}{sum_{i=1}^{n} a_i}][b = frac{sum_{i=1}^{n} b_i x_i}{sum_{i=1}^{n} b_i}]So, ( a ) and ( b ) are both linear combinations of the ( x_i )s with positive coefficients. Since all ( a_i ) and ( b_i ) are positive integers, the denominators are positive, ensuring that ( a ) and ( b ) are well-defined.Now, let's consider the difference ( |a - b| ). Maybe I can express this difference in terms of the ( x_i )s and the weights ( a_i ) and ( b_i ).Calculating ( a - b ):[a - b = frac{sum_{i=1}^{n} a_i x_i}{sum_{i=1}^{n} a_i} - frac{sum_{i=1}^{n} b_i x_i}{sum_{i=1}^{n} b_i}]To combine these two fractions, I can find a common denominator, which would be ( left( sum_{i=1}^{n} a_i right) left( sum_{i=1}^{n} b_i right) ). So,[a - b = frac{left( sum_{i=1}^{n} a_i x_i right) left( sum_{j=1}^{n} b_j right) - left( sum_{i=1}^{n} b_i x_i right) left( sum_{j=1}^{n} a_j right)}{left( sum_{i=1}^{n} a_i right) left( sum_{j=1}^{n} b_j right)}]This looks a bit complicated, but maybe I can simplify it. Let's expand the numerator:[left( sum_{i=1}^{n} a_i x_i right) left( sum_{j=1}^{n} b_j right) = sum_{i=1}^{n} sum_{j=1}^{n} a_i b_j x_i][left( sum_{i=1}^{n} b_i x_i right) left( sum_{j=1}^{n} a_j right) = sum_{i=1}^{n} sum_{j=1}^{n} a_j b_i x_i]Subtracting these two:[sum_{i=1}^{n} sum_{j=1}^{n} a_i b_j x_i - sum_{i=1}^{n} sum_{j=1}^{n} a_j b_i x_i = sum_{i=1}^{n} sum_{j=1}^{n} (a_i b_j - a_j b_i) x_i]Hmm, that's still a bit messy. Maybe there's another approach.Let me think about the relationship between ( a ) and ( b ). Since both are weighted averages of the same set of ( x_i )s, their difference should somehow relate to how the weights ( a_i ) and ( b_i ) differ.Perhaps I can express ( a - b ) as a weighted sum of the differences ( x_i - x_j ). But I'm not sure if that's the right direction.Wait, another idea: maybe I can use the concept of convex combinations or something similar. Since ( a ) and ( b ) are both convex combinations (because the weights sum to 1 after normalization), they lie within the convex hull of the ( x_i )s.But I'm not sure if that directly helps with the inequalities given.Let me try to consider the first inequality ( |a - b| leq |a - x_p| ). This suggests that the distance between ( a ) and ( b ) is less than or equal to the distance from ( a ) to some ( x_p ). So, ( a ) is closer to ( b ) than it is to ( x_p ).Similarly, the second inequality ( |a - x_p| leq |x_q - x_p| ) suggests that the distance from ( a ) to ( x_p ) is less than or equal to the distance between two ( x )s, specifically ( x_q ) and ( x_p ).Putting these together, it seems that ( a ) is somewhere between ( b ) and ( x_p ), and ( x_p ) is not too far from ( x_q ).Maybe I can use the triangle inequality here. Let's recall that for any real numbers, ( |a - b| leq |a - x_p| + |x_p - b| ). But I'm not sure if that's directly applicable.Alternatively, perhaps I can consider the maximum and minimum values among the ( x_i )s. Let me denote ( x_{text{min}} = min{x_1, x_2, ldots, x_n} ) and ( x_{text{max}} = max{x_1, x_2, ldots, x_n} ). Then, ( a ) and ( b ) must lie between ( x_{text{min}} ) and ( x_{text{max}} ) because they are weighted averages.So, ( x_{text{min}} leq a, b leq x_{text{max}} ). Therefore, the maximum possible distance between any two ( x_i )s is ( |x_{text{max}} - x_{text{min}}| ).But how does this relate to ( |a - b| ) and ( |a - x_p| )?Maybe I can argue that ( |a - b| ) is less than or equal to the maximum distance between any two ( x_i )s, which is ( |x_{text{max}} - x_{text{min}}| ). But I need to connect this to ( |a - x_p| ).Wait, perhaps if I consider the distances from ( a ) to each ( x_i ), there must be some ( x_p ) such that ( |a - x_p| ) is at least ( |a - b| ). Because ( b ) is another weighted average, it can't be farther from ( a ) than some ( x_i ) is.Similarly, ( |a - x_p| ) can't be larger than the maximum distance between any two ( x_i )s, because ( a ) is a weighted average and thus lies within the range of the ( x_i )s.So, putting it all together, I can say that ( |a - b| leq |a - x_p| leq |x_q - x_p| ) for some ( x_p ) and ( x_q ).But I need to formalize this reasoning.Let me try to write it out step by step.First, since ( a ) is a weighted average, it lies within the convex hull of the ( x_i )s. Similarly, ( b ) also lies within the same convex hull. Therefore, the distance between ( a ) and ( b ) is bounded by the maximum distance between any two points in the convex hull, which is the maximum distance between any two ( x_i )s.So, ( |a - b| leq |x_q - x_p| ) for some ( x_p ) and ( x_q ).But I need to show that ( |a - b| leq |a - x_p| leq |x_q - x_p| ).Hmm, perhaps I can consider the maximum deviation of ( a ) from the ( x_i )s. Let me define ( D = max{ |a - x_i| : i = 1, 2, ldots, n } ). Then, ( |a - b| leq D ) because ( b ) is another weighted average, so it can't be farther from ( a ) than the farthest ( x_i ) is.Therefore, ( |a - b| leq D ). But ( D = |a - x_p| ) for some ( x_p ). So, ( |a - b| leq |a - x_p| ).Now, since ( |a - x_p| ) is the maximum deviation, it must be less than or equal to the maximum distance between any two ( x_i )s, which is ( |x_q - x_p| ) for some ( x_q ).Thus, ( |a - x_p| leq |x_q - x_p| ).Putting these together, we have ( |a - b| leq |a - x_p| leq |x_q - x_p| ), which is what we needed to prove.Wait, does this make sense? Let me double-check.1. ( a ) is a weighted average, so it lies within the range of the ( x_i )s.2. ( b ) is another weighted average, so it also lies within the same range.3. The distance between ( a ) and ( b ) can't be greater than the maximum distance between any two ( x_i )s.4. The maximum distance from ( a ) to any ( x_i ) is at least the distance between ( a ) and ( b ).5. Therefore, there exists an ( x_p ) such that ( |a - x_p| ) is at least ( |a - b| ).6. Additionally, ( |a - x_p| ) can't exceed the maximum distance between any two ( x_i )s, so there exists an ( x_q ) such that ( |x_q - x_p| ) is at least ( |a - x_p| ).Yes, this seems to hold. So, by considering the maximum deviations and the properties of weighted averages, we can establish the required inequalities.I think this approach is solid. I just need to make sure I present it clearly in the final proof.

💡Okay, so I have this problem with vectors AB, AC, and BC. The vectors are given as AB = (m, 5), AC = (4, n), and BC = (7, 6). I need to find the value of m + n. Hmm, let me think about how to approach this.First, I remember that vectors can be added and subtracted using their components. Maybe I can use the relationship between these vectors to set up some equations. I think the key here is to use the fact that vector BC can be expressed in terms of vectors AB and AC.Wait, how exactly? I recall that in vector addition, if you have points A, B, and C, then vector BC is equal to vector AC minus vector AB. Is that right? Let me visualize it. If I start at point A and go to point B, that's vector AB. Then from B to C is vector BC. Alternatively, from A to C is vector AC. So, vector AC should be equal to vector AB plus vector BC. Hmm, so maybe vector BC is vector AC minus vector AB. Yeah, that makes sense because if you go from A to C, it's like going from A to B and then from B to C.So, mathematically, that would be:[overrightarrow{BC} = overrightarrow{AC} - overrightarrow{AB}]Plugging in the given vectors:[(7, 6) = (4, n) - (m, 5)]Now, I can break this down into its components. For the x-component:[7 = 4 - m]And for the y-component:[6 = n - 5]Okay, so I have two equations here. Let's solve for m and n.Starting with the x-component equation:[7 = 4 - m]Subtract 4 from both sides:[7 - 4 = -m][3 = -m]Multiply both sides by -1:[m = -3]Alright, so m is -3. Now, let's solve for n using the y-component equation:[6 = n - 5]Add 5 to both sides:[6 + 5 = n][11 = n]So, n is 11. Now, the question asks for m + n. Let's add them together:[m + n = -3 + 11 = 8]Wait, that seems straightforward. Let me double-check my steps to make sure I didn't make any mistakes.First, I used the relationship between the vectors correctly. Vector BC is indeed vector AC minus vector AB. Then, I broke it down into components and solved for m and n. The calculations seem correct: 4 - (-3) is 7, and 11 - 5 is 6, which matches the given vector BC. So, everything checks out.I think I'm confident with this answer. m is -3, n is 11, so m + n is 8.

💡First, I need to determine the increase in the stock price by subtracting the opening price from the closing price.Next, I'll calculate the percentage increase by dividing the price increase by the opening price and then multiplying by 100 to convert it to a percentage.Finally, I'll present the calculated percentage increase as the answer.

💡Okay, so I have this problem about an arithmetic sequence. Let me see. It says that the sequence has a positive common difference, which means each term is bigger than the one before it. The first three terms add up to 15, and their product is 80. I need to find the sum of the 11th, 12th, and 13th terms. Hmm, okay.First, let me recall what an arithmetic sequence is. It's a sequence where each term after the first is obtained by adding a constant difference. So, if the first term is ( a_1 ), then the second term is ( a_1 + d ), the third term is ( a_1 + 2d ), and so on. Here, ( d ) is the common difference, which is positive.Given that ( a_1 + a_2 + a_3 = 15 ). Let me write that out in terms of ( a_1 ) and ( d ). So, ( a_1 + (a_1 + d) + (a_1 + 2d) = 15 ). Simplifying that, it becomes ( 3a_1 + 3d = 15 ). If I factor out the 3, it's ( 3(a_1 + d) = 15 ). Dividing both sides by 3, I get ( a_1 + d = 5 ). So, that's one equation.Now, the product of the first three terms is 80. So, ( a_1 times a_2 times a_3 = 80 ). Substituting the terms, that's ( a_1 times (a_1 + d) times (a_1 + 2d) = 80 ). Hmm, that seems a bit more complicated. Maybe I can use the first equation to substitute into this one.From the first equation, I have ( a_1 + d = 5 ), so ( a_2 = 5 ). That might be useful. So, ( a_2 = 5 ). Then, the product equation becomes ( a_1 times 5 times a_3 = 80 ). So, ( 5a_1a_3 = 80 ). Dividing both sides by 5, I get ( a_1a_3 = 16 ).But I also know that ( a_3 = a_1 + 2d ). And since ( a_1 + d = 5 ), I can express ( d = 5 - a_1 ). So, substituting back into ( a_3 ), we have ( a_3 = a_1 + 2(5 - a_1) ). Let me compute that: ( a_1 + 10 - 2a_1 = -a_1 + 10 ). So, ( a_3 = -a_1 + 10 ).Now, I have ( a_1a_3 = 16 ), and ( a_3 = -a_1 + 10 ). So, substituting, ( a_1(-a_1 + 10) = 16 ). Let's write that out: ( -a_1^2 + 10a_1 = 16 ). Bringing all terms to one side, ( -a_1^2 + 10a_1 - 16 = 0 ). Multiplying both sides by -1 to make it a bit nicer: ( a_1^2 - 10a_1 + 16 = 0 ).Now, I have a quadratic equation: ( a_1^2 - 10a_1 + 16 = 0 ). Let me try to factor this. Looking for two numbers that multiply to 16 and add up to -10. Hmm, 16 can be factored as 2 and 8, or 4 and 4. 2 and 8 add up to 10. Since the middle term is -10a, the factors should both be negative. So, (a - 2)(a - 8) = 0. Let me check: ( (a - 2)(a - 8) = a^2 - 10a + 16 ). Yes, that's correct.So, the solutions are ( a_1 = 2 ) or ( a_1 = 8 ). Hmm, but since the common difference is positive, let's see what that implies. If ( a_1 = 2 ), then ( d = 5 - a_1 = 3 ). So, the sequence would be 2, 5, 8, 11, 14, etc. That makes sense because each term is increasing by 3.If ( a_1 = 8 ), then ( d = 5 - 8 = -3 ). But the problem says the common difference is positive, so this solution is invalid. Therefore, ( a_1 = 2 ) and ( d = 3 ).Okay, so now I know the first term is 2 and the common difference is 3. The question asks for ( a_{11} + a_{12} + a_{13} ). Let's figure out what those terms are.In an arithmetic sequence, the nth term is given by ( a_n = a_1 + (n - 1)d ). So, let's compute each term:- ( a_{11} = 2 + (11 - 1) times 3 = 2 + 10 times 3 = 2 + 30 = 32 )- ( a_{12} = 2 + (12 - 1) times 3 = 2 + 11 times 3 = 2 + 33 = 35 )- ( a_{13} = 2 + (13 - 1) times 3 = 2 + 12 times 3 = 2 + 36 = 38 )Now, let's add them up: ( 32 + 35 + 38 ). Let me compute that step by step. 32 + 35 is 67, and 67 + 38 is 105. So, the sum of the 11th, 12th, and 13th terms is 105.Wait, just to double-check, maybe there's a smarter way to compute this without calculating each term individually. Since it's an arithmetic sequence, the sum of three consecutive terms can be expressed as ( 3 times ) the middle term. So, ( a_{11} + a_{12} + a_{13} = 3a_{12} ). Since ( a_{12} = 35 ), multiplying by 3 gives 105. Yep, that's the same result. So, that confirms it.Therefore, the answer should be 105, which is option B.**Final Answer**boxed{B}