# In which of the following situations,

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 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.

Solution:

(i) $t_{n}$ denotes the taxi fare (in Rs.) for the first $n \mathrm{~km}$.

Now, $t_{1}=15$,

$\mathrm{t}_{2}=15+8=23$

$\mathrm{t}_{3}=23+8=31$

$\mathrm{t}_{4}=31+8=39, \ldots$

List of fares after $1 \mathrm{~km}, 2 \mathrm{~km}, 3 \mathrm{~km}, 4 \mathrm{~km}, \ldots$ respectively is $15,23,31,39, \ldots$ (in Rs.)

Here, $t_{2}-t_{1}=t_{3}-t_{2}=t_{4}-t_{3}=\ldots .=8$.

Thus, the list forms an AP.

(ii) $\mathrm{t}_{1}=\mathrm{x}$ units $; \mathrm{t}_{2}=\mathrm{x}-\frac{1}{4} \mathrm{x}=\frac{3}{4} \mathrm{x}$ units;

$t_{3}=\frac{3}{4} x-\frac{1}{4}\left(\frac{3}{4} x\right)=\frac{3}{4} x-\frac{3}{16} x=\frac{9}{16} x$ units

$t_{4}=\frac{9}{16} x-\frac{1}{4}\left(\frac{9}{16} x\right)=\frac{27}{64} x$ units

The list of numbers is $x, \frac{3}{4} x, \frac{9}{16} x, \frac{27}{64} x, \ldots$

It is not an $A P$ because $t_{2}-t_{1} \neq t_{3}-t_{2}$.

(iii) Cost of digging for first metre $=150$

Cost of digging for first 2 metres

= 150 + 50 = 200

Cost of digging for first 3 metres

= 200 + 50 = 250

Cost of digging for first 4 metres

= 250 + 50 = 300

Clearly, $150,200,250,300 \ldots$ forms an A.P.

Here, $\mathrm{t}_{2}-\mathrm{t}_{1}=\mathrm{t}_{3}-\mathrm{t}_{2}=\mathrm{t}_{4}-\mathrm{t}_{3}=\ldots . .=50$.

Thus, the list forms an AP.

(iv) We know that if Rs P is deposited at $\mathrm{r} \%$ compound interest per annum for $\mathrm{n}$ years, our money will be $P\left(1+\frac{r}{100}\right)^{n}$ after $n$ years.

Therefore, after every year, our money will be

$10000\left(1+\frac{8}{100}\right), 10000\left(1+\frac{8}{100}\right)^{2}$

$10000\left(1+\frac{8}{100}\right)^{3}, 10000\left(1+\frac{8}{100}\right)^{4}, \ldots .$

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.