2. Өрнектін мәнін табыңыз: \( \frac{a+2 b}{b} \), егер \( \frac{a}{b}=2 \). 3. Есептеңіз \( \operatorname{tg} \alpha \), егер \( \sin \alpha=\frac{5}{13}, \frac{\pi}{2}<\alpha<\pi \). 4. Ара қашықтығы 900 км екі каладан бір-біріне қарама-қарсы екі п тада кездескен. Бірінші пойыз екіншіден 1 сағ кеш шығып, екінші ылдамдығынан 5 км/сағ артык жылдамдыкпен жүрген. Әр пойызд быныз. ;. Геометриялык прогрессияда \( \mathrm{n}=6 \mathrm{q}=0,5 \mathrm{~b}_{\mathrm{n}}=3 \) екендігі белгілі. Пр. үшесін, алғашкы п мүшесінің косындысын табыңыз.

Alright, I need to solve these mathematical problems step by step. Let's tackle them one by one. **Problem 1:** Find the value of the expression \( \frac{a + 2b}{b} \) given that \( \frac{a}{b} = 2 \). First, I'll substitute the given ratio into the expression. Since \( \frac{a}{b} = 2 \), it means that \( a = 2b \). Plugging this into the expression: \[ \frac{a + 2b}{b} = \frac{2b + 2b}{b} = \frac{4b}{b} = 4 \] So, the value of the expression is 4. **Problem 2:** Find \( \operatorname{tg} \alpha \) given that \( \sin \alpha = \frac{5}{13} \) and \( \frac{\pi}{2} < \alpha < \pi \). First, I recall that \( \operatorname{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} \). I already have \( \sin \alpha = \frac{5}{13} \). To find \( \cos \alpha \), I'll use the Pythagorean identity: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \] Plugging in the known value: \[ \left(\frac{5}{13}\right)^2 + \cos^2 \alpha = 1 \Rightarrow \frac{25}{169} + \cos^2 \alpha = 1 \Rightarrow \cos^2 \alpha = 1 - \frac{25}{169} = \frac{144}{169} \] Since \( \alpha \) is in the second quadrant where cosine is negative: \[ \cos \alpha = -\frac{12}{13} \] Now, calculating \( \operatorname{tg} \alpha \): \[ \operatorname{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{5}{13}}{-\frac{12}{13}} = -\frac{5}{12} \] So, \( \operatorname{tg} \alpha = -\frac{5}{12} \). **Problem 3:** Two trains are 900 km apart and moving towards each other. The first train leaves 1 hour before the second and travels at 5 km/h faster than the second train, which travels at 5 km/h. Find the speeds of both trains. Let's denote the speed of the second train as \( v \) km/h. Then, the speed of the first train is \( v + 5 \) km/h. Since the first train leaves 1 hour earlier, it travels for \( t + 1 \) hours, and the second train travels for \( t \) hours. The distance covered by the first train is: \[ \text{Distance}_1 = (v + 5)(t + 1) \] The distance covered by the second train is: \[ \text{Distance}_2 = v \times t \] Since the trains are moving towards each other, the sum of the distances they cover equals 900 km: \[ (v + 5)(t + 1) + v \times t = 900 \] Expanding the equation: \[ v t + 5 t + v + 5 + v t = 900 \Rightarrow 2 v t + 5 t + v + 5 = 900 \] This equation has two variables, \( v \) and \( t \), which makes it difficult to solve directly. To find a solution, I'll assume that the trains meet after a certain time \( t \). However, without additional information, there are infinitely many solutions depending on the value of \( t \). Therefore, more information is needed to determine the exact speeds of the trains. **Problem 4:** In a geometric progression, \( n = 6 \), \( q = 0.5 \), and \( b_n = 3 \). Find the third term and the sum of the first \( n \) terms. First, recall that the \( n \)-th term of a geometric progression is given by: \[ b_n = b_1 \times q^{n-1} \] Given \( b_n = 3 \), \( q = 0.5 \), and \( n = 6 \): \[ 3 = b_1 \times (0.5)^{5} \Rightarrow b_1 = \frac{3}{(0.5)^5} = \frac{3}{0.03125} = 96 \] So, the first term \( b_1 = 96 \). Next, the third term \( b_3 \) is: \[ b_3 = b_1 \times q^{2} = 96 \times (0.5)^2 = 96 \times 0.25 = 24 \] Now, the sum of the first \( n \) terms \( S_n \) in a geometric progression is: \[ S_n = b_1 \times \frac{1 - q^n}{1 - q} \] Plugging in the values: \[ S_6 = 96 \times \frac{1 - (0.5)^6}{1 - 0.5} = 96 \times \frac{1 - 0.015625}{0.5} = 96 \times \frac{0.984375}{0.5} = 96 \times 1.96875 = 189 \] So, the third term is 24, and the sum of the first 6 terms is 189.