Let the first term be a and the common ratio be r. Then, 2nd term = ar = 12 and 4th term = ar^3 = 48. Dividing these gives r^2 = 4, so r = 2.

Let the first term be a and the common ratio be r. Then, 2nd term = ar = 8 and 4th term = ar^3 = 32. Dividing gives r^2 = 4, so r = 2.

Let the first term be a and the common difference be d. From the equations a + d = 10 and a + 4d = 16, we can solve for d to find it is 2.

Let the first term be a and the common difference be d. From a + d = 10 and a + 4d = 16, we can find a + 2d = 12, which is the 3rd term.

Let the first term be a and the common difference be d. From the equations a + d = 15 and a + 3d = 25, we can find d = 5.

Let the first term be a and the common difference be d. From the equations a + d = 8 and a + 3d = 14, we can find a = 6.

Let the first term be a and the common difference be d. From the equations a + d = 8 and a + 4d = 14, we can find the 3rd term a + 2d = 10.

Let the first term be a and the common difference be d. From the equations a + d = 8 and a + 4d = 20, we can solve for a to find it equals 4.

The 3rd term is given by ar^2. So, 12 = a(2^2) => a = 12/4 = 3.

Let the first term be a. Then, the 3rd term is ar^2 = 27. Thus, a * 3^2 = 27, giving a = 3.

Let the first term be a and the common difference be d. We have a + 2d = 12 and a + 6d = 24. Subtracting these gives 4d = 12, so d = 3.

Let the first term be a and the common difference be d. From the equations a + 2d = 15 and a + 5d = 24, solving gives d = 3.

Let the first term be a and the common difference be d. From the equations a + 2d = 15 and a + 6d = 27, solving gives d = 3.

Let the first term be a and the common difference be d. Then, a + 2d = 12 and a + 6d = 24. Solving these gives d = 3.

Let the first term be a and the common difference be d. From the equations a + 4d = 15 and a + 9d = 30, we can find d = 3.

Let the first term be a and the common difference be d. From the equations a + 4d = 20 and a + 9d = 35, we can solve for a to find it is 10.

Let the first term be a and the common difference be d. From the equations a + 5d = 30 and a + 8d = 45, we can find d = 5.

Using the formula for the nth term, a + 6d = 25. Substituting d = 3 gives a + 18 = 25, thus a = 7.

Using the formula for the nth term, a + 6d = 50. Substituting d = 5 gives a + 30 = 50, hence a = 20.

According to the Arrhenius equation, an increase in activation energy results in a decrease in the rate constant k.

Using the formula A = P(1 + r)^t, we have 1210 = 1000(1 + r)^2. Solving gives r = 0.1 or 10%.

Using the formula A(t) = A_0 e^(-ζω_nt), we find ζ = 0.2.

Maximum velocity V_max = Aω. If A is doubled, V_max also doubles.

The total energy E in SHM is given by E = (1/2)kA². If A is doubled, E becomes (1/2)k(2A)² = 4(1/2)kA², which is quadrupled.

Total energy (E) in SHM is proportional to the square of the amplitude (A). If A is doubled, E becomes 4A^2, hence it quadruples.

Maximum velocity V_max = ωA. If A is halved, V_max is also halved.

The total energy in simple harmonic motion is proportional to the square of the amplitude. If amplitude is doubled, energy increases by a factor of 2^2 = 4.

The total energy of a simple harmonic oscillator is proportional to the square of the amplitude. If the amplitude is doubled, the energy increases by a factor of 2^2 = 4.

The total energy in SHM is proportional to the square of the amplitude. If amplitude is halved, energy is reduced to (1/2)^2 = 1/4, which is halved.

The total energy (E) in a simple harmonic oscillator is given by E = (1/2)kA². If amplitude (A) increases, energy increases.