💡First, I recognize that the sequence is arithmetic, meaning the difference between consecutive terms is constant.Given the first term ( a_1 = 1 ) and the second term ( a_2 = 3 ), I can find the common difference ( d ) by subtracting ( a_1 ) from ( a_2 ):[ d = a_2 - a_1 = 3 - 1 = 2 ]Next, I use the general formula for the ( n )-th term of an arithmetic sequence:[ a_n = a_1 + (n - 1)d ]Substituting the known values:[ a_n = 1 + (n - 1) times 2 = 2n - 1 ]To find the term number ( n ) where ( a_n = 217 ), I set up the equation:[ 2n - 1 = 217 ]Solving for ( n ):[ 2n = 218 ][ n = 109 ]Therefore, the number 217 is the 109th term in the sequence.

💡Okay, so I have this problem where I need to calculate the product P, which is given by multiplying a series of fractions starting from 3/2 up to 30/29. The expression looks like this:P = (3/2) × (4/3) × (5/4) × ... × (29/28) × (30/29)At first glance, this seems like a lot of fractions multiplied together, and I'm not sure how to approach it. Maybe I can write out the first few terms to see if there's a pattern or something that cancels out.Let me write them out:(3/2) × (4/3) × (5/4) × (6/5) × ... × (29/28) × (30/29)Hmm, I notice that in each fraction, the numerator is one more than the denominator. So, for example, 3 is one more than 2, 4 is one more than 3, and so on.Now, if I look at the numerators and denominators, I see that the 3 in the numerator of the first fraction cancels with the 3 in the denominator of the second fraction. Similarly, the 4 in the numerator of the second fraction cancels with the 4 in the denominator of the third fraction. This pattern continues all the way through the sequence.So, if I keep canceling these terms, most of them will disappear. Let's see:- The 3 in the numerator of (3/2) cancels with the 3 in the denominator of (4/3).- The 4 in the numerator of (4/3) cancels with the 4 in the denominator of (5/4).- This continues until the 29 in the numerator of (29/28) cancels with the 29 in the denominator of (30/29).After all these cancellations, what's left is just the first denominator, which is 2, and the last numerator, which is 30. So, the entire product simplifies to 30 divided by 2.Calculating that gives me 15.Wait, let me double-check to make sure I didn't miss anything. If I have 30 terms starting from 3/2 up to 30/29, does that mean all intermediate terms cancel out? Yes, because each numerator cancels with the denominator of the next fraction. So, the only terms that remain are the first denominator and the last numerator.Therefore, the product P should indeed be 30/2, which simplifies to 15.I think that makes sense. So, the final answer is 15.

💡Okay, so I have this problem here: Find the least integer x such that cos(x) equals sin(x²), where x is measured in degrees and x is at least 1. The options are from 8 to 12, so I need to figure out which one is the smallest integer that satisfies this equation.First, I remember that cos and sin are related through some trigonometric identities. One of them is that sin(θ) is equal to cos(90° - θ). So maybe I can use that to rewrite the equation.Let me write down the equation again: cos(x) = sin(x²). Using the identity, sin(x²) is equal to cos(90° - x²). So substituting that in, the equation becomes cos(x) = cos(90° - x²).Now, when do two cosine functions equal each other? I recall that cos(A) = cos(B) if and only if A = B + 360°k or A = -B + 360°k for some integer k. So, applying that here, we have two cases:1. x = 90° - x² + 360°k2. x = -(90° - x²) + 360°kLet me write these out more clearly.**Case 1:** x = 90° - x² + 360°k**Case 2:** x = -90° + x² + 360°kNow, since we're looking for the smallest integer x that's at least 1, I should probably start with the smallest possible k, which is k = 0, because higher k values would give larger x, which isn't what we want.**Starting with Case 1:**x = 90° - x²Let me rearrange this equation:x² + x - 90 = 0This is a quadratic equation in terms of x. I can solve this using the quadratic formula: x = [-b ± sqrt(b² - 4ac)] / (2a)Here, a = 1, b = 1, c = -90.Plugging in the values:x = [-1 ± sqrt(1 + 360)] / 2x = [-1 ± sqrt(361)] / 2x = [-1 ± 19] / 2So, two solutions:1. x = (-1 + 19) / 2 = 18 / 2 = 92. x = (-1 - 19) / 2 = -20 / 2 = -10Since x has to be at least 1, we discard -10. So, x = 9 is a solution from Case 1.**Now, let's look at Case 2:**x = -90° + x²Rearranging:x² - x - 90 = 0Again, using the quadratic formula:x = [1 ± sqrt(1 + 360)] / 2x = [1 ± sqrt(361)] / 2x = [1 ± 19] / 2So, two solutions:1. x = (1 + 19) / 2 = 20 / 2 = 102. x = (1 - 19) / 2 = -18 / 2 = -9Again, x has to be at least 1, so we discard -9. So, x = 10 is a solution from Case 2.Now, comparing the solutions from both cases, we have x = 9 and x = 10. Since we're looking for the least integer x, 9 is smaller than 10. So, x = 9 is the answer.But wait, just to make sure, maybe I should plug these values back into the original equation to verify.**Testing x = 9:**cos(9°) ≈ 0.9877sin(9²) = sin(81°) ≈ 0.9877Yes, they are approximately equal.**Testing x = 10:**cos(10°) ≈ 0.9848sin(10²) = sin(100°) ≈ 0.9848They are also approximately equal.So both x = 9 and x = 10 satisfy the equation, but since we need the smallest integer, 9 is the answer.I think that's solid. I don't see any mistakes in my reasoning. I used the trigonometric identity correctly, solved the quadratics, checked the solutions, and verified by plugging them back in. So, yeah, x = 9 is the least integer that satisfies the equation.**Final Answer**boxed{B}

💡Okay, so I have this problem here: Given that the polynomial ( x^2 - kx + 24 ) has only positive integer roots, I need to find the average of all distinct possibilities for ( k ). Hmm, let me try to figure this out step by step.First, I remember that for a quadratic polynomial of the form ( x^2 - (p + q)x + pq ), the roots are ( p ) and ( q ). So, in this case, comparing it to the given polynomial ( x^2 - kx + 24 ), I can say that:1. The product of the roots ( p times q = 24 ).2. The sum of the roots ( p + q = k ).So, my goal is to find all pairs of positive integers ( (p, q) ) such that their product is 24. Once I have all these pairs, I can calculate ( k ) for each pair by adding the two numbers, and then find the average of all these ( k ) values.Alright, let me list out all the positive integer pairs ( (p, q) ) where ( p times q = 24 ). I'll start by finding the factors of 24.24 can be factored as:- 1 × 24- 2 × 12- 3 × 8- 4 × 6Wait, are there more? Let me check:- 5 doesn't divide 24 evenly, so 5 × something isn't a factor pair.- 6 × 4 is already listed.- 7 doesn't divide 24 either.- 8 × 3 is already there.- 9, 10, 11, etc., don't divide 24.So, I think those are all the positive integer pairs: (1,24), (2,12), (3,8), (4,6). Now, for each of these pairs, I'll calculate ( k = p + q ):- For (1,24): ( k = 1 + 24 = 25 )- For (2,12): ( k = 2 + 12 = 14 )- For (3,8): ( k = 3 + 8 = 11 )- For (4,6): ( k = 4 + 6 = 10 )So, the possible values of ( k ) are 25, 14, 11, and 10. Now, I need to find the average of these ( k ) values. To do that, I'll add them all up and then divide by the number of values.First, let's add them:25 + 14 = 3939 + 11 = 5050 + 10 = 60So, the total sum is 60. There are 4 distinct values of ( k ). Therefore, the average is ( frac{60}{4} = 15 ).Wait a second, let me double-check my addition to be sure:25 + 14 is indeed 39.39 + 11 is 50.50 + 10 is 60. Yep, that's correct.And dividing 60 by 4 gives 15. So, the average of all distinct possibilities for ( k ) is 15.I think that's it. I considered all positive integer pairs that multiply to 24, calculated the corresponding ( k ) values, summed them up, and found the average. Everything seems to check out.**Final Answer**The average of all distinct possibilities for ( k ) is boxed{15}.

💡Alright, so I have this problem here: I need to expand the cube of (√3 x - 1), which is written as (√3 x - 1)^3, and express it in the form a0 + a1 x + a2 x² + a3 x³. Then, I have to find the value of (a0 + a2)² - (a1 + a3)². Hmm, okay, let me think about how to approach this.First, I remember that expanding a binomial like (a + b)^3 can be done using the binomial theorem. The binomial theorem says that (a + b)^n is the sum from k=0 to n of (n choose k) a^(n-k) b^k. So, in this case, n is 3, a is √3 x, and b is -1. Maybe I can use that to expand it.Let me write that out step by step. So, (√3 x - 1)^3. Using the binomial theorem, that would be:C(3,0)*(√3 x)^3*(-1)^0 + C(3,1)*(√3 x)^2*(-1)^1 + C(3,2)*(√3 x)^1*(-1)^2 + C(3,3)*(√3 x)^0*(-1)^3.Let me compute each term one by one.First term: C(3,0) is 1, (√3 x)^3 is (√3)^3 x³, which is 3√3 x³, and (-1)^0 is 1. So, the first term is 3√3 x³.Second term: C(3,1) is 3, (√3 x)^2 is (√3)^2 x², which is 3 x², and (-1)^1 is -1. So, the second term is 3*3 x²*(-1) = 9 x²*(-1) = -9 x².Third term: C(3,2) is 3, (√3 x)^1 is √3 x, and (-1)^2 is 1. So, the third term is 3*√3 x*1 = 3√3 x.Fourth term: C(3,3) is 1, (√3 x)^0 is 1, and (-1)^3 is -1. So, the fourth term is 1*1*(-1) = -1.Putting it all together, the expansion is:3√3 x³ - 9 x² + 3√3 x - 1.So, comparing this to a0 + a1 x + a2 x² + a3 x³, we can identify the coefficients:a0 is the constant term, which is -1.a1 is the coefficient of x, which is 3√3.a2 is the coefficient of x², which is -9.a3 is the coefficient of x³, which is 3√3.Okay, so now I have a0 = -1, a1 = 3√3, a2 = -9, a3 = 3√3.Now, the problem asks for (a0 + a2)² - (a1 + a3)². Let me compute each part step by step.First, compute a0 + a2. That's (-1) + (-9) = -10.Then, compute a1 + a3. That's 3√3 + 3√3 = 6√3.Now, compute (a0 + a2)², which is (-10)² = 100.Next, compute (a1 + a3)², which is (6√3)². Let's compute that: (6√3)² = 6²*(√3)² = 36*3 = 108.So, now, subtract these two: (a0 + a2)² - (a1 + a3)² = 100 - 108 = -8.Wait, that seems straightforward. But let me double-check my steps to make sure I didn't make any mistakes.First, expanding (√3 x - 1)^3 using the binomial theorem. Let me verify each term again.First term: C(3,0)*(√3 x)^3*(-1)^0 = 1*(3√3 x³)*1 = 3√3 x³. Correct.Second term: C(3,1)*(√3 x)^2*(-1)^1 = 3*(3 x²)*(-1) = -9 x². Correct.Third term: C(3,2)*(√3 x)^1*(-1)^2 = 3*(√3 x)*1 = 3√3 x. Correct.Fourth term: C(3,3)*(√3 x)^0*(-1)^3 = 1*1*(-1) = -1. Correct.So, the expansion is indeed 3√3 x³ - 9 x² + 3√3 x - 1. So, the coefficients are correct.Then, a0 = -1, a1 = 3√3, a2 = -9, a3 = 3√3. Correct.Then, a0 + a2 = -1 + (-9) = -10. Correct.a1 + a3 = 3√3 + 3√3 = 6√3. Correct.(a0 + a2)² = (-10)² = 100. Correct.(a1 + a3)² = (6√3)² = 36*3 = 108. Correct.Subtracting: 100 - 108 = -8. Correct.So, the value is -8. Hmm, that seems right. But just to be thorough, maybe there's another way to approach this problem without expanding the entire expression. Sometimes, using substitution or properties of polynomials can help.I remember that for a polynomial P(x), P(1) gives the sum of the coefficients, and P(-1) gives the alternating sum of the coefficients. Maybe that can help here.Let me define P(x) = (√3 x - 1)^3. Then, P(1) = (√3*1 - 1)^3 = (√3 - 1)^3. Similarly, P(-1) = (√3*(-1) - 1)^3 = (-√3 - 1)^3.Now, P(1) = a0 + a1 + a2 + a3.P(-1) = a0 - a1 + a2 - a3.So, if I compute P(1) and P(-1), I can get two equations:1. P(1) = a0 + a1 + a2 + a3 = (√3 - 1)^3.2. P(-1) = a0 - a1 + a2 - a3 = (-√3 - 1)^3.Now, the problem asks for (a0 + a2)^2 - (a1 + a3)^2.Notice that (a0 + a2)^2 - (a1 + a3)^2 can be factored as (a0 + a2 - a1 - a3)(a0 + a2 + a1 + a3).Wait, that's interesting. Because (a0 + a2)^2 - (a1 + a3)^2 = (a0 + a2 - a1 - a3)(a0 + a2 + a1 + a3).Looking back at the expressions for P(1) and P(-1):P(1) = a0 + a1 + a2 + a3.P(-1) = a0 - a1 + a2 - a3.If I add P(1) and P(-1), I get:P(1) + P(-1) = 2(a0 + a2).Similarly, if I subtract P(-1) from P(1), I get:P(1) - P(-1) = 2(a1 + a3).So, that gives me:a0 + a2 = [P(1) + P(-1)] / 2.a1 + a3 = [P(1) - P(-1)] / 2.Therefore, (a0 + a2)^2 - (a1 + a3)^2 = ([P(1) + P(-1)] / 2)^2 - ([P(1) - P(-1)] / 2)^2.Let me compute that.First, let me compute P(1) and P(-1):P(1) = (√3 - 1)^3.P(-1) = (-√3 - 1)^3.Let me compute P(1) first:(√3 - 1)^3. Let's expand this.Using the binomial theorem again:(a - b)^3 = a^3 - 3a^2 b + 3a b^2 - b^3.So, (√3)^3 - 3*(√3)^2*(1) + 3*(√3)*(1)^2 - (1)^3.Compute each term:(√3)^3 = (√3)*(√3)^2 = √3*3 = 3√3.-3*(√3)^2*1 = -3*3*1 = -9.3*(√3)*1^2 = 3√3.-1^3 = -1.So, adding them up: 3√3 - 9 + 3√3 - 1 = (3√3 + 3√3) + (-9 -1) = 6√3 - 10.So, P(1) = 6√3 - 10.Similarly, compute P(-1) = (-√3 - 1)^3.Again, using the binomial theorem:(a + b)^3 = a^3 + 3a^2 b + 3a b^2 + b^3.Here, a = -√3, b = -1.So, (-√3)^3 + 3*(-√3)^2*(-1) + 3*(-√3)*(-1)^2 + (-1)^3.Compute each term:(-√3)^3 = - (√3)^3 = -3√3.3*(-√3)^2*(-1) = 3*(3)*(-1) = 9*(-1) = -9.3*(-√3)*(-1)^2 = 3*(-√3)*(1) = -3√3.(-1)^3 = -1.Adding them up: -3√3 - 9 - 3√3 -1 = (-3√3 - 3√3) + (-9 -1) = -6√3 - 10.So, P(-1) = -6√3 - 10.Now, let's compute a0 + a2 and a1 + a3.a0 + a2 = [P(1) + P(-1)] / 2 = [ (6√3 -10) + (-6√3 -10) ] / 2.Compute numerator: 6√3 -10 -6√3 -10 = (6√3 -6√3) + (-10 -10) = 0 -20 = -20.So, a0 + a2 = (-20)/2 = -10.Similarly, a1 + a3 = [P(1) - P(-1)] / 2 = [ (6√3 -10) - (-6√3 -10) ] / 2.Compute numerator: 6√3 -10 +6√3 +10 = (6√3 +6√3) + (-10 +10) = 12√3 + 0 = 12√3.So, a1 + a3 = (12√3)/2 = 6√3.Therefore, (a0 + a2)^2 - (a1 + a3)^2 = (-10)^2 - (6√3)^2 = 100 - (36*3) = 100 - 108 = -8.So, same result as before. That's reassuring.Alternatively, I could have noticed that (a0 + a2)^2 - (a1 + a3)^2 is equal to (P(1) + P(-1))/2 squared minus (P(1) - P(-1))/2 squared, which simplifies to [P(1) + P(-1)]²/4 - [P(1) - P(-1)]²/4.This can be factored as [ (P(1) + P(-1))² - (P(1) - P(-1))² ] / 4.Using the identity A² - B² = (A - B)(A + B), where A = P(1) + P(-1) and B = P(1) - P(-1).So, [ (P(1) + P(-1) - (P(1) - P(-1))) (P(1) + P(-1) + P(1) - P(-1)) ] / 4.Simplify the terms inside:First term: P(1) + P(-1) - P(1) + P(-1) = 2 P(-1).Second term: P(1) + P(-1) + P(1) - P(-1) = 2 P(1).So, the numerator becomes 2 P(-1) * 2 P(1) = 4 P(1) P(-1).Therefore, the entire expression is [4 P(1) P(-1)] / 4 = P(1) P(-1).So, (a0 + a2)^2 - (a1 + a3)^2 = P(1) P(-1).Therefore, instead of computing a0 + a2 and a1 + a3 separately, I could have just multiplied P(1) and P(-1).Let me compute that:P(1) = (√3 -1)^3 = 6√3 -10.P(-1) = (-√3 -1)^3 = -6√3 -10.So, P(1) * P(-1) = (6√3 -10)(-6√3 -10).Let me compute this product.Multiply the two binomials:(6√3)(-6√3) + (6√3)(-10) + (-10)(-6√3) + (-10)(-10).Compute each term:First term: 6√3 * -6√3 = -36*(√3)^2 = -36*3 = -108.Second term: 6√3 * -10 = -60√3.Third term: -10 * -6√3 = 60√3.Fourth term: -10 * -10 = 100.Now, add them all together:-108 -60√3 +60√3 +100.Simplify:-108 +100 + (-60√3 +60√3) = (-8) + 0 = -8.So, P(1)*P(-1) = -8.Therefore, (a0 + a2)^2 - (a1 + a3)^2 = -8.This is the same result as before. So, regardless of the method, I get -8.I think this is solid. I expanded the polynomial, found the coefficients, computed the required expression, and then verified using properties of polynomials and substitution. Both methods led me to the same answer, so I feel confident that -8 is correct.Just to recap, the key steps were:1. Expand (√3 x -1)^3 using the binomial theorem to find the coefficients a0, a1, a2, a3.2. Compute a0 + a2 and a1 + a3.3. Calculate (a0 + a2)^2 - (a1 + a3)^2.Alternatively, recognizing that this expression can be related to P(1) and P(-1), and using properties of polynomials to find the result without explicitly expanding.Either way, the answer is -8.**Final Answer**The value of (a_{0}+a_{2})^{2}-(a_{1}+a_{3})^{2} is boxed{-8}.

💡First, I need to recall the formula for the area of a trapezium, which is ( text{Area} = frac{1}{2} times (text{sum of the lengths of the two parallel sides}) times text{height} ).In this problem, the cross-section of the tunnel is a trapezium with a top width of 15 meters and a bottom width of 5 meters. The area of this trapezium is given as 400 square meters. I need to find the depth of the tunnel, which corresponds to the height of the trapezium.Using the formula, I can set up the equation:[400 = frac{1}{2} times (15 + 5) times h]Simplifying the equation:[400 = frac{1}{2} times 20 times h][400 = 10 times h]Solving for ( h ):[h = frac{400}{10} = 40 text{ meters}]