In which of the following situations,

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.