Write first four terms of the AP,

Solution:

(i) $\mathrm{t}_{1}=\mathrm{a}=10$,

$t_{2}=10+d=10+10=20$

$t_{3}=20+d=20+10=30$

$\mathrm{t}_{4}=30+\mathrm{d}=30+10=40, \ldots .$

Thus, the AP is $10,20,30,40, \ldots$

(ii) Given $\mathrm{a}=-2$ and $\mathrm{d}=0$

$\mathrm{t}_{1}=-2, \mathrm{t}_{2}=-2+0=-2$

$t_{3}=-2+0=-2, t_{4}=-2+0=-2, \ldots .$

Thus, the AP is $-2,-2,-2,-2, \ldots$

(iii) $\mathrm{a}=4, \mathrm{~d}=-3$

$\mathrm{t}_{1}=\mathrm{a}=4$

$\mathrm{t}_{2}=\mathrm{a}_{1}+\mathrm{d}=4-3=1$

$\mathrm{t}_{3}=\mathrm{a}_{2}+\mathrm{d}=1-3=-2$

$\mathrm{t}_{4}=\mathrm{a}_{3}+\mathrm{d}=-2-3=-5$

Therefore, the series will be $4,1,-2-5 \ldots$

First four terms of this A.P. will be $4,1,-2$ and $-5$.

(iv) $\mathrm{a}=-1, \mathrm{~d}=\frac{1}{2}$

$\mathrm{t}_{1}=\mathrm{a}=-1$

$\mathrm{t}_{2}=\mathrm{a}_{1}+\mathrm{d}=-1+\frac{1}{2}=-\frac{1}{2}$

$\mathrm{t}_{3}=\mathrm{a}_{2}+\mathrm{d}=-\frac{1}{2}+\frac{1}{2}=0$

Clearly, the series will be

$-1,-\frac{1}{2}, 0, \frac{1}{2} \cdots \cdots \cdots \cdots$

First four terms of this A.P. will be

$-1,-\frac{1}{2}, 0$ and $-\frac{1}{2}$

(v) $\mathrm{a}=-1.25, \mathrm{~d}=-0.25$

$t_{1}=a=-1.25$

$\mathrm{t}_{2}=\mathrm{a}_{1}+\mathrm{d}=-1.25-0.25=-1.50$

$\mathrm{t}_{3}=\mathrm{a}_{2}+\mathrm{d}=-1.50-0.25=-1.75$

$\mathrm{t}_{4}=\mathrm{a}_{3}+\mathrm{d}=-1.75-0.25=-2.00$

Clearly, the series will be $-1.25,-1.50,-1.75$, $\begin{array}{lllll}-2.00 & \ldots & \ldots & \ldots .\end{array}$

First four terms of this A.P. will be $-1.25$,

$-1.50,-1.75$ and $-2.00$