Question

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each $km$ when the fare is $Rs.15$ for the first $km$ and $Rs.8$ for each additional $km$.
(ii) The amount of air present in a cylinder when a vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs $Rs.150$ for the first metre and rises by $Rs.50$ for each subsequent metre.
(iv) The amount of money in the account every year, when $Rs.10000$ is deposited at compound interest at $8\%$ per annum.

Answer 1

(i) Fare for first km $=$ $Rs.15$

Fare for second km $=$ $Rs.(15+8)=$ $Rs.23$

Fare for third km $=$ $Rs.(23+8)=Rs.31$

Here, each subsequent term is obtained by adding a fixed number $\left(8\right)$ to the previous term.

Hence, it is in A.P.

(ii) Let us assume, initial quantity of air $=1\dots \dots \left(i\right)$

Therefore, quantity removed in first step $=\frac{1}{4}$

Remaining quantity after first step $=1-\frac{1}{4}=\frac{3}{4}\dots \dots \left(ii\right)$

Quantity removed in second step $=$ $\frac{3}{4}\times \frac{1}{4}=\frac{3}{16}$

Remaining quantity after second step $=$ $\frac{3}{4}-\frac{3}{16}=\frac{9}{16}\dots \dots \left(iii\right)$

Here, each subsequent term is not obtained by adding a fixed number to the previous term.

Hence, it is not an AP.

(iii) Cost of digging of $1^{st}$ meter $=150$

Cost of digging of $2^{nd}$ meter $=150+50=200$

Cost of digging of $3^{rd}$ meter $=200+50=250$

Here, each subsequent term is obtained by adding a fixed number $\left(50\right)$ to the previous term.

Hence, it is an AP.

(iv) Amount in the beginning $=$ Rs. $10000$

Interest at the end of $1^{st}$ year @ $8\%$ = $10000\times 8\%=800$

Thus, amount at the end of $1^{st}$ year $=10000+800=10800$

Interest at the end of $2^{nd}$ year @ $8\%$ = $10800\times 8\%=864$

Thus, amount at the end of $2^{nd}$ year $=10800+864=11664$

Since, each subsequent term is not obtained by adding a fixed number to the previous term.

Hence, it is not an AP.