Write the arithmetic progression when first term a and common difference d are as follows:

Solution:

In the given problem, we are given its first term (a) and common difference (d).

We need to find the A.P

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

Now, as $a=4$

A.P would be represented by $a, a_{1}, a_{2}, a_{3}, a_{4}, \ldots \ldots$

So,

$a_{1}=a+d$

$a_{1}=4+(-3)$

 

$a_{1}=1$

Similarly,

$a_{2}=a_{1}+d$

$a_{2}=1+(-3)$

 

$a_{2}=-2$

Also,

$a_{3}=a_{2}+d$

$a_{3}=-2+(-3)$

 

$a_{3}=-5$

Further,

$a_{4}=a_{3}+d$

$a_{4}=-5+(-3)$

 

$a_{4}=-8$

Therefore, A.P with $a=4$ and $d=-3$ is $4,1,-2,-5,-8, \ldots$

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

Now, as $a=-1$

A.P would be represented by 

So,

$a_{1}=a+d$

$a_{1}=-1+\left(\frac{1}{2}\right)$

$a_{1}=\frac{-2+1}{2}$

$a_{1}=\frac{-1}{2}$

Similarly,

$a_{2}=a_{1}+d$

$a_{2}=\frac{-1}{2}+\left(\frac{1}{2}\right)$

$a_{2}=0$

Also,

$a_{3}=a_{2}+d$

$a_{3}=0+\left(\frac{1}{2}\right)$

$a_{3}=\frac{1}{2}$

Further,

$a_{4}=a_{3}+d$

$a_{4}=\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)$

$a_{4}=\frac{2}{2}$

$a_{4}=1$

Therefore, A.P with $a=4$ and $d=-3$ is $-1, \frac{-1}{2}, 0, \frac{1}{2}, 1, \ldots . .$

(iii) $a=-1.5, d=-0.5$

Now, as $a=-1.5$

A.P would be represented by $a, a_{1}, a_{2}, a_{3}, a_{4}, \ldots . .$

 

So,

$a_{1}=a+d$

$a_{1}=-1.5+(-0.5)$

 

$a_{1}=-2$

Similarly,

$a_{2}=a_{1}+d$

$a_{2}=-2+(-0.5)$

$a_{2}=-2.5$

Also,

$a_{3}=a_{2}+d$

$a_{3}=-2.5+(-0.5)$

 

$a_{3}=-3$

Further,

$a_{4}=a_{3}+d$

$a_{4}=-3+(-0.5)$

 

$a_{4}=-3.5$

Therefore, A.P with $-1.5,-2,-2.5,-3,-3.5, \ldots$