(i) The cost of digging a well for the first metre is Rs $150$ and rises by Rs $20$ for each coming after metre.
(ii) The amount of air present in the cylinder when a vacuum pump removes each time $1/4$ of their remaining in the cylinder. 

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:

(i) Given,

Cost of digging a well for the first meter $\left( {{c}_{1}} \right)=Rs.150.$

And, the cost rises by $Rs.20$ for each succeeding meter

Then,

Cost of digging for the second meter $\left( {{c}_{2}} \right)=Rs.150+Rs20=Rs170$

Cost of digging for the third meter $\left( {{c}_{3}} \right)=Rs.170+Rs20=Rs210$

Hence, its clearly seen that the costs of digging a well for different lengths are $150,170,190,210,$ ….

Evidently, this series is in A∙P.

With first term $\left( a \right)=150$, common difference $\left( d \right)=20$

(ii) Given,

Let the initial volume of air in a cylinder be $V$ liters each time ${{3}^{th}}/4$ of air in a remaining i.e

$1-1/4$

First time, the air in cylinder is $V$.

Second time, the air in cylinder is $3/4$ $V.$

Third time, the air in cylinder is ${{\left( 3/4 \right)}^{2}}$ $V.$

Thus, series is $V,3/4V,{{\left( 3/4 \right)}^{2}}V,{{\left( 3/4 \right)}^{2}}V,$ ….

Hence, the above series is not a A.P.