answer:(1) Let the first term of the arithmetic sequence { a_{n} } be a_{1} and the common difference be d. From the given conditions, we have the system of equations: begin{cases} a_{1} + d = 0 2a_{1} + 12d = -10 end{cases} Solving this system, we obtain a_{1} = 1 and d = -1. Therefore, the general term formula for the arithmetic sequence { a_{n} } is: a_{n} = 2 - n. (2) Let S_{n} denote the sum of the first n terms of the sequence left{ frac{a_{n}}{2^{n-1}} right}. Then, we have: S_{n} = 1 + frac{0}{2} + frac{-1}{2^{2}} + cdots + frac{2-n}{2^{n-1}}. Dividing both sides by 2, we obtain: frac{S_{n}}{2} = frac{1}{2} + frac{0}{2^{2}} + frac{-1}{2^{3}} + cdots + frac{2-n}{2^{n}}. Subtracting the two equations, we get: frac{S_{n}}{2} = 1 - left( frac{1}{2} + frac{1}{2^{2}} + frac{1}{2^{3}} + cdots + frac{1}{2^{n-1}} right) - frac{2-n}{2^{n}}. The expression in the parentheses is a geometric series with the first term frac{1}{2} and the common ratio frac{1}{2}. Its sum can be found using the formula: S_{n} = frac{a_{1}(1 - r^{n})}{1 - r}, where a_{1} is the first term, r is the common ratio, and n is the number of terms. In this case, we have a_{1} = frac{1}{2}, r = frac{1}{2}, and n = n-1. Plugging these values into the formula, we get: 1 - left( frac{1}{2} + frac{1}{2^{2}} + frac{1}{2^{3}} + cdots + frac{1}{2^{n-1}} right) = 1 - frac{frac{1}{2}left(1 - left(frac{1}{2}right)^{n-1}right)}{1 - frac{1}{2}} = frac{1}{2^{n-1}}. So, frac{S_{n}}{2} = 1 - frac{1}{2^{n-1}} - frac{2-n}{2^{n}} = frac{n}{2^{n}}. Thus, the sum of the first n terms of the sequence left{ frac{a_{n}}{2^{n-1}} right} is: boxed{S_{n} = frac{n}{2^{n-1}}}.

answer:Applying the Triangle Inequality, we have [|z - 8| + |z - 7i| = |z - 8| + |7i - z| ge |(z - 8) + (7i - z)| = |-8 + 7i|.] Calculating the magnitude of (-8 + 7i), [ |-8 + 7i| = sqrt{(-8)^2 + 7^2} = sqrt{64 + 49} = sqrt{113}. ] Since we are given that |z - 8| + |z - 7i| = 17, and sqrt{113} leq 17, equality holds if z lies on the line segment connecting 8 and 7i. To minimize |z|, we locate z at the perpendicular projection of the origin onto the line connecting 8 and 7i. Defining points ( A = (8, 0) ) and ( B = (0, 7) ) in the complex plane, the line equation is ( y = -frac{7}{8}x + 7 ). Setting x = y = 0 and using the point to line distance formula, [ text{Distance} = frac{|-7 cdot 0 - 8 cdot 0 + 7|}{sqrt{(-7)^2 + 8^2}} = frac{7}{sqrt{113}}. ] Thus, the smallest possible value of |z| is boxed{frac{7}{sqrt{113}}}.

answer:1. Let's denote the sides of the right-angled triangle by ( a ), ( b ), and ( c ), where ( c ) is the hypotenuse. 2. According to the Pythagorean Theorem for a right-angled triangle: [ c^2 = a^2 + b^2 ] 3. We are given that the sum of the squares of all three side lengths is 1800: [ a^2 + b^2 + c^2 = 1800 ] 4. Substituting ( a^2 + b^2 ) from the Pythagorean Theorem into the given equation: [ c^2 + c^2 = 1800 ] 5. Simplifying the equation, we find: [ 2c^2 = 1800 ] 6. Solving for ( c^2 ): [ c^2 = frac{1800}{2} implies c^2 = 900 ] 7. Taking the square root of both sides to find ( c ): [ c = sqrt{900} ] 8. Since side lengths are positive, we find: [ c = 30 ] # Conclusion: [ boxed{30} ]

answer:When n geq 2, we have a_n = S_n - S_{n-1} =(2^{n}-1)-(2^{n-1}-1)=2^{n-1}. When n=1, we have a_1=S_1=1, which is consistent with the above formula. So, a_n=2^{n-1}, and therefore a_3=2^{2}=4. The sum of the first 4 terms is S_4= frac{1 times (1-2^{4})}{1-2}=15. Hence, frac{S_4}{a_3}= frac{15}{4}. Thus, the answer is boxed{frac{15}{4}}. This problem involves identifying the sequence as a geometric sequence and then finding a_3 and S_4 to determine their ratio. It tests the knowledge of the sum formula for geometric sequences and is of medium difficulty.