Write first four terms of the AP, when the first term $a$ and the common difference $d$ are given as follows:
(i) $a = 10, d = 10$
(ii) $a = -2, d = 0$
(iii) $a = 4, d = -3$
(iv) $a = -1, d = \frac{1}{2}$
(v) $a = -1.25, d = -0.25$

To do:

We have to write the AP, when the first term $a$ and the common difference $d$ are given.

Solution:

(i) First term $a_1=a=10$

Second term $a_2=a_1+d=10+(10)=20$

Third term $a_3=a_2+d=20+(10)=30$

Fourth term $a_4=a_3+d=30+(10)=40$

Therefore, the first four terms of the AP are $10, 20, 30, 40$.

(ii) First term $a_1=a=-2$

Second term $a_2=a_1+d=-2+0=-2$

Third term $a_3=a_2+d=-2+0=-2$

Fourth term $a_4=a_3+d=-2+0=-2$

Therefore, the first four terms of the AP are $-2, -2, -2, -2$. 

(iii) First term $a_1=a=4$

Second term $a_2=a_1+d=4+(-3)=4-3=1$

Third term $a_3=a_2+d=1+(-3)=1-3=-2$

Fourth term $a_4=a_3+d=-2+(-3)=-2-3=-5$

Therefore, the first four terms of the AP are $4, 1, -2, -5$.

(iv) First term $a_1=a=-1$

Second term $a_2=a_1+d=-1+\frac{1}{2}=\frac{-1\times2+1}{2}=\frac{-1}{2}$

Third term $a_3=a_2+d=\frac{-1}{2}+\frac{1}{2}=0$

Fourth term $a_4=a_3+d=0+\frac{1}{2}=\frac{1}{2}$

Therefore, the first four terms of the given AP are $-1, \frac{-1}{2}, 0, \frac{1}{2}$. 

(v) First term $a_1=a=-1.25$

Second term $a_2=a_1+d=-1.25+(-0.25)=-1.25-0.25=-1.50$

Third term $a_3=a_2+d=-1.50+(-0.25)=-1.50-0.25=-1.75$

Fourth term $a_4=a_3+d=-1.75+(-0.25)=-1.75-0.25=-2$

Therefore, the first four terms of the given AP are $-1.25, -1.50, -1.75, -2$.