(i) $a=4,d=-3$

(ii) $a=-1,d={1}/{2}\;$

An arithmetic progression is a number’s sequence such that the difference between the consecutive terms is constant.

Formula for this is: $an=d\left( n-1 \right)+c,$

Solution:

We know that, if first term $\left( a \right)=a$ and common difference $=d,$ then the arithmetic series is:$a,a+d,a+2d,a+3d$

(i) $a=4,d=-3$

Given, first term $\left( a \right)=4$

Common difference $\left( d \right)=-3$

Then arithmetic progression is: $a,a+d,a+2d,a+3d$ ……

⇒ $4,4-3,4+2\left( -3 \right),4+3\left( -3 \right),$ ……

⇒ $4,1,-2,-5,-8$……..

(ii) $a=-1,d=1/2$

Given, first term $\left( a \right)=-1$

Common difference $\left( d \right)=1/2$

Then arithmetic progression is: $a,a+d,a+2d,a+3d,$

⇒ $-1,-1+1/2+2{\scriptstyle{}^{1}/{}_{2}},-1+3{\scriptstyle{}^{1}/{}_{2,}}$ …

⇒ $-1,-1/2,0,1/2$