Quarter of one figure is equal to three times of second figure, If 12 is added in the four times of second figure, the resultant figure is equal to three times of first figure. Write down both the figures in the sequence?

Let the first figure be \( x \) and the second figure be \( y \).

From the problem, we can derive two equations:

1. Quarter of the first figure is equal to three times the second figure:
\[
\frac{x}{4} = 3y
\]
Therefore, multiplying both sides by 4:
\[
x = 12y \quad \text{(Equation 1)}
\]

2. If 12 is added to four times the second figure, the resultant figure equals three times the first figure:
\[
4y + 12 = 3x
\]
Substituting \( x \) from Equation 1:
\[
4y + 12 = 3(12y)
\]
Simplifying this gives:
\[
4y + 12 = 36y
\]
Rearranging the equation gives:
\[
12 = 36y - 4y
\]
\[
12 = 32y
\]
Dividing by 32:
\[
y = \frac{12}{32} = \frac{3}{8}
\]

Now, substituting \( y \) back into Equation 1 to find \( x \):
\[
x = 12y = 12 \times \frac{3}{8} = \frac{36}{8} = \frac{9}{2} = 4.5
\]

Thus, the first figure \( x \) is \( 4.5 \) and the second figure \( y \) is \( \frac{3}{8} \).

To summarize:
- First figure \( x = 4.5 \)
- Second figure \( y = \frac{3}{8} \)

In sequence: \( 4.5, \frac{3}{8} \).