answer:To find the length of the train, we need to calculate the total distance covered by the train while crossing the bridge. This distance is the sum of the length of the train and the length of the bridge. First, we convert the speed of the train from kilometers per hour (kmph) to meters per second (m/s) to match the units of the bridge length and the time taken. Speed in m/s = Speed in kmph × (1000 meters / 1 kilometer) × (1 hour / 3600 seconds) Speed in m/s = 54 × (1000 / 1) × (1 / 3600) Speed in m/s = 54 × 1000 / 3600 Speed in m/s = 54000 / 3600 Speed in m/s = 15 m/s Now, we use the formula for distance, which is the product of speed and time: Distance = Speed × Time The time taken is given as 58.9952803775698 seconds, and we have calculated the speed to be 15 m/s. So the total distance covered while crossing the bridge is: Distance = 15 m/s × 58.9952803775698 s Distance ≈ 15 × 58.9952803775698 Distance ≈ 884.929205663547 m This distance is the sum of the length of the train and the length of the bridge. We know the length of the bridge is 720 meters, so we can find the length of the train (TrainLength) by subtracting the bridge length from the total distance: TrainLength = Total Distance - Bridge Length TrainLength ≈ 884.929205663547 m - 720 m TrainLength ≈ 164.929205663547 m Therefore, the length of the train is approximately boxed{164.93} meters.

answer:The problem involves analyzing a parabola defined by ( y^2 = 2px ) where ( p > 0 ), and determining certain properties about the line segment ( AB ) between two points on the parabola. Part 1: Given points ( A ) and ( B ) on the parabola ( y^2 = 2px ), with midpoint ( M(a, t) ): 1. Let ( A ) have coordinates ( (x_1, y_1) ) and ( B ) have coordinates ( (x_2, y_2) ). Since ( M ) is the midpoint of ( AB ): [ 2a = x_1 + x_2, quad 2t = y_1 + y_2 ] 2. As ( A ) and ( B ) lie on the parabola: [ y_1^2 = 2px_1, quad y_2^2 = 2px_2 ] 3. When ( t neq 0 ), let the slope of ( AB ) be ( k ) (and ( k neq 0 )): [ k = frac{y_1 - y_2}{x_1 - x_2} = frac{2p(y_1 - y_2)}{y_1^2 - y_2^2} = frac{2p}{y_1 + y_2} = frac{p}{t} ] The equation of the vertical bisector line ( l ) of segment ( AB ) is: [ y - t = -frac{1}{k}(x - a) = -frac{t}{p}(x - a) ] which simplifies to: [ p cdot y = t(p + a - x) ] When ( x = p + a ), ( y = 0 ). Therefore, the vertical bisector line passes through the point ( N(p + a, 0) ). 4. When ( t = 0 ), points ( A ) and ( B ) are symmetric about the x-axis, and the vertical bisector is the x-axis itself, which also passes through the point ( N(p + a, 0) ). In both cases, the vertical bisector line ( l ) always passes through ( N(p + a, 0) ). Therefore, [ bIacksquare ] Part 2: To find the maximum area of ( triangle ANB ) given ( |FN| = 2a ) where ( a > 1 ): 1. ( F ) is the focus of the parabola ( y^2 = 2px ), located at ( Fleft(frac{p}{2}, 0right) ). Given ( |FN| = 2a ), we have: [ left| p + a - frac{p}{2} right| = 2a ] Solving for ( p ): [ p = 2a ] Therefore, the equation of the parabola is: [ y^2 = 4ax ] 2. Consider when ( t neq 0 ), the equation of the line ( AB ) is: [ 2ax - ty + t^2 - 2a^2 = 0 ] Eliminating ( x ) and simplifying, we obtain: [ y^2 - 2ty + 2(t^2 - 2a^2) = 0 ] The roots ( y_1 ) and ( y_2 ) of this quadratic equation satisfy: [ y_1 + y_2 = 2t, quad y_1 cdot y_2 = 2(t^2 - 2a^2) ] where the discriminant should be positive for real roots: [ Delta = 4t^2 - 4 cdot 2(t^2 - 2a^2) = 4(4a^2 - t^2) > 0 Rightarrow -2a < t < 2a ] 3. Calculating distance ( |AB| ): [ |AB|^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 = left( left(frac{t}{2a}right)^2 + 1 right)(y_1 - y_2)^2 ] [ |AB|^2 = frac{1}{4a^2}(t^2 + 4a^2)(4a^2 - t^2) ] [ |AB| = frac{1}{2} sqrt{(t^2 + 4a^2)(4a^2 - t^2)} ] 4. Distance from point ( N ) to line ( AB ): [ d = frac{|2a(p + a) + t^2 - 2a^2|}{sqrt{2a^2 + t^2}} = sqrt{4a^2 + t^2} ] 5. Formula for area ( S_{triangle ANB} ): [ S_{triangle ANB} = frac{1}{2} |AB| cdot d = frac{1}{2a}(4a^2 + t^2) sqrt{4a^2 - t^2} ] Considering all possible values of ( t ) (i.e., ( t in [-2a, 2a] )) and using the Arithmetic Mean-Geometric Mean (AM-GM) inequality for three positive numbers, we find: [ f(t) = frac{1}{2a} left( frac{1}{2}(4a^2 + t^2)(8a^2 - 2t^2) right)^{frac{3}{2}} ] [ leq frac{sqrt{2}}{4a} left( frac{16a^2}{3} right)^{frac{3}{2}} = frac{16sqrt{6}}{9} a^2 ] The equality holds when ( t = pm frac{2}{3} sqrt{3a} ). Therefore, the maximum area of ( triangle ANB ) is: [ boxed{frac{16sqrt{6}}{9} a^2} ]

answer:When n=1, we have S_1 = 2a_1 - 2. Therefore, a_1 = 2. When n=2, we have S_2 = 2a_2 - 2. Therefore, a_2 = a_1 + 2 = 4. Hence, the answer is boxed{4}. **Analysis:** By using S_n = 2a_n - 2 and taking n as 1 and 2, we can find the value of a_2.

answer:1. **Initial Setup:** We are given two conditions: [ 2m mid 3n - 2 quad text{and} quad 2n mid 3m - 2 ] We need to find all ordered pairs of positive integers ((m, n)) that satisfy these conditions. 2. **Substitution:** Let (m = 2m_1) and (n = 2n_1), where (m_1) and (n_1) are positive integers. Substituting these into the given conditions, we get: [ 2(2m_1) mid 3(2n_1) - 2 quad text{and} quad 2(2n_1) mid 3(2m_1) - 2 ] Simplifying, we obtain: [ 4m_1 mid 6n_1 - 2 quad text{and} quad 4n_1 mid 6m_1 - 2 ] 3. **Simplification:** Dividing both sides of each condition by 2, we get: [ 2m_1 mid 3n_1 - 1 quad text{and} quad 2n_1 mid 3m_1 - 1 ] 4. **Oddness Condition:** For (2m_1) to divide (3n_1 - 1) and (2n_1) to divide (3m_1 - 1), (m_1) and (n_1) must be odd. This is because an even number cannot divide an odd number unless the odd number is zero, which is not possible here. 5. **Combining Conditions:** We now have: [ 2m_1 mid 3n_1 - 1 quad text{and} quad 2n_1 mid 3m_1 - 1 ] This implies: [ 4m_1n_1 mid (3n_1 - 1)(3m_1 - 1) ] Expanding the product on the right-hand side: [ (3n_1 - 1)(3m_1 - 1) = 9m_1n_1 - 3m_1 - 3n_1 + 1 ] Therefore: [ 4m_1n_1 mid 9m_1n_1 - 3m_1 - 3n_1 + 1 ] 6. **Divisibility Condition:** Simplifying the divisibility condition: [ 4m_1n_1 mid 9m_1n_1 - 3m_1 - 3n_1 + 1 ] This implies: [ m_1n_1 mid 3(m_1 + n_1) - 1 ] 7. **Bounding (n_1):** Without loss of generality, assume (m_1 geq n_1). Then: [ m_1n_1 leq 3(m_1 + n_1) - 1 ] Since (m_1 geq n_1), we have: [ m_1n_1 leq 6m_1 - 1 ] Dividing both sides by (m_1): [ n_1 leq 6 - frac{1}{m_1} ] Since (n_1) is an odd positive integer, the possible values for (n_1) are 1, 3, and 5. 8. **Checking Possible Values:** - For (n_1 = 1): [ 2m_1 mid 2 implies m_1 = 1 ] Thus, ((m, n) = (2, 2)). - For (n_1 = 3): [ 2m_1 mid 8 implies m_1 = 1, 2, 4 ] Since (m_1) must be odd, (m_1 = 1) is the only possibility, but (m_1 geq n_1) is not satisfied. - For (n_1 = 5): [ 2m_1 mid 14 implies m_1 = 1, 7 ] Since (m_1) must be odd, (m_1 = 7) is the only possibility. Thus, ((m, n) = (14, 10)) and ((m, n) = (10, 14)). Conclusion: The only pairs ((m, n)) that satisfy the given conditions are ((2, 2)), ((10, 14)), and ((14, 10)). The final answer is (boxed{(2, 2), (10, 14), (14, 10)}).