Identify which of the following situations form an arithmetic progression (AP): (i) Taxi fare for 1 km is ₹15 and for each additional km is ₹8. (ii) Amount of air in a cylinder when vacuum pump remove

The fare for the first km is ₹ 15. For each additional km, the fare is ₹ 8.
The list of fares for 1 km, 2 km, 3 km, ... is:
$a_1 = 15$
$a_2 = 15 + 8 = 23$
$a_3 = 23 + 8 = 31$
$a_4 = 31 + 8 = 39$
The sequence is $15, 23, 31, 39, \dots$
The common difference is $d = 23 - 15 = 8$, $31 - 23 = 8$, $39 - 31 = 8$.
Since the common difference is constant, this situation forms an AP.

Let the initial amount of air in the cylinder be $V$. A vacuum pump removes $\frac{1}{4}$ of the air remaining at a time.
After the first removal: $V_1 = V - \frac{1}{4}V = \frac{3}{4}V$
After the second removal: $V_2 = V_1 - \frac{1}{4}V_1 = \frac{3}{4}V_1 = \frac{3}{4}\left(\frac{3}{4}V\right) = \frac{9}{16}V$
After the third removal: $V_3 = V_2 - \frac{1}{4}V_2 = \frac{3}{4}V_2 = \frac{3}{4}\left(\frac{9}{16}V\right) = \frac{27}{64}V$
The sequence of air remaining is $V, \frac{3}{4}V, \frac{9}{16}V, \frac{27}{64}V, \dots$
The differences between consecutive terms are:
$d_1 = \frac{3}{4}V - V = -\frac{1}{4}V$
$d_2 = \frac{9}{16}V - \frac{3}{4}V = \frac{9}{16}V - \frac{12}{16}V = -\frac{3}{16}V$
Since $d_1 \neq d_2$, the common difference is not constant. This situation does not form an AP.

The cost of digging the first metre is ₹ 150. The cost rises by ₹ 50 for each subsequent metre.
The list of costs for 1 m, 2 m, 3 m, ... is:
$C_1 = 150$
$C_2 = 150 + 50 = 200$
$C_3 = 200 + 50 = 250$
$C_4 = 250 + 50 = 300$
The sequence is $150, 200, 250, 300, \dots$
The common difference is $d = 200 - 150 = 50$, $250 - 200 = 50$, $300 - 250 = 50$.
Since the common difference is constant, this situation forms an AP.

Initial deposit = ₹ 10000. Compound interest at 8% per annum.
The amount after $n$ years is given by the formula $A = P(1 + r)^n$.
$A_0 = 10000$
$A_1 = 10000(1 + 0.08)^1 = 10000(1.08) = 10800$
$A_2 = 10000(1.08)^2 = 10000(1.1664) = 11664$
$A_3 = 10000(1.08)^3 = 10000(1.259712) = 12597.12$
The sequence of amounts is $10000, 10800, 11664, 12597.12, \dots$
The differences between consecutive terms are:
$d_1 = 10800 - 10000 = 800$
$d_2 = 11664 - 10800 = 864$
Since $d_1 \neq d_2$, the common difference is not constant. This situation does not form an AP.

The situations that form an arithmetic progression are:

*i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km. This forms an AP because the fare increases by a constant amount (₹ 8) for each additional km.

*iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre. This forms an AP because the cost increases by a constant amount (₹ 50) for each additional metre.