1 1~6 11 f) ) u_(1)+u_(3)+u_(5)=-12 u_(1)u_(2)u_(3)=8 1) u_(9)=5u_(2) và u_(13)=2u_(6)+5

The provided examples show how to solve for the first term (u1) and common difference (d) or common ratio (q) of arithmetic and geometric sequences, respectively, given various conditions. Let's tackle the problems you've presented:**f) **This problem deals with an arithmetic sequence. We know that:* un = u1 + (n-1)d (where un is the nth term, u1 is the first term, and d is the common difference)Let's rewrite the equations using this formula:* u1 + (u1 + 2d) + (u1 + 4d) = -12 => 3u1 + 6d = -12 => u1 + 2d = -4 (Equation 1)* u1(u1 + d)(u1 + 2d) = 8 (Equation 2)Substitute Equation 1 into Equation 2:* u1(u1 + d)(-4) = 8 => u1(u1 + d) = -2This gives us a system of two equations with two unknowns:* u1 + 2d = -4* u1(u1 + d) = -2Solving this system requires substitution or elimination. Let's solve for d in Equation 1: d = (-4 - u1)/2Substitute this into the second equation:u1(u1 + (-4 - u1)/2) = -2Solving this quadratic equation for u1 will give you the first term. Once you have u1, you can easily find d using Equation 1. This will likely involve some trial and error or using the quadratic formula.**i) u9 = 5u2 and u13 = 2u6 + 5**This problem deals with an arithmetic sequence. Using the formula un = u1 + (n-1)d:* u9 = u1 + 8d = 5(u1 + d) => u1 + 8d = 5u1 + 5d => 4u1 - 3d = 0 (Equation 1)* u13 = u1 + 12d = 2(u1 + 5d) + 5 => u1 + 12d = 2u1 + 10d + 5 => u1 - 2d = -5 (Equation 2)Now we have a system of two linear equations:* 4u1 - 3d = 0* u1 - 2d = -5Solve this system (e.g., using substitution or elimination) to find u1 and d. The easiest approach here is to solve Equation 2 for u1 (u1 = 2d - 5) and substitute into Equation 1.Remember to check your solutions by plugging them back into the original equations to verify they are correct. The solutions will be integers based on the examples provided.