1. The problem asks to find the first four terms of an arithmetic progression (A.P.) where the first term $a = -1$ and the common difference $d = \frac{1}{2}$. 2. Recall the formula for the $n$th term of an A.P.: $$a_n = a + (n-1)d$$ 3. To find the second term $a_2$, we use: $$a_2 = a + d = -1 + \frac{1}{2}$$ 4. To add $-1$ and $\frac{1}{2}$, express $-1$ as a fraction with denominator 2: $$-1 = \frac{-2}{2}$$ 5. Now add the fractions: $$\frac{-2}{2} + \frac{1}{2} = \frac{-2 + 1}{2} = \frac{-1}{2}$$ 6. So, the second term is $-\frac{1}{2}$. 7. Similarly, the third term is: $$a_3 = a + 2d = -1 + 2 \times \frac{1}{2} = -1 + 1 = 0$$ 8. The fourth term is: $$a_4 = a + 3d = -1 + 3 \times \frac{1}{2} = -1 + \frac{3}{2} = \frac{-2}{2} + \frac{3}{2} = \frac{1}{2}$$ 9. Therefore, the first four terms of the A.P. are: $$-1, -\frac{1}{2}, 0, \frac{1}{2}$$