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# 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.

To do:

We have to find whether the given situations are possible.

Solution:

(i) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.

Let

$a_1 = Rs.\ 15$

This implies,

$a_2 = Rs.\ 15 + Rs.\ 8 = Rs.\ 23$

$a_3 = Rs.\ 23 + Rs.\ 8 = Rs.\ 31$

Now, the list of fares is $Rs.\ 15, Rs.\ 23, Rs.\ 31$

$a_2 - a_1 = Rs.\ 23 - Rs.\ 15$

$= Rs.\ 8$

$a_3 - a_2 = Rs.\ 31 - Rs.\ 23$

$= Rs.\ 8$

Here,

$a_2 - a_1 = a_3 - a_2$

Therefore, the given list of fares forms an A.P.

(ii) The amount of air present in the cylinder when a vacuum pump removes each time $\frac{1}{4}$ of the remaining in the cylinder.

Let the air present n the cylinder be $1$. The amount of air removed the first time $=1\times\frac{1}{4}=\frac{1}{4}$

The amount of air remaining after removing the air first time $=1-\frac{1}{4}=\frac{1\times4-1}{4}=\frac{3}{4}$

The amount of air removed the second time $=\frac{3}{4}\times\frac{1}{4}=\frac{3}{16}$

The amount of air remaining after removing the air second time $=\frac{3}{4}-\frac{3}{16}=\frac{4\times3-3}{16}=\frac{9}{16}$

The amount of air removed the third time $=\frac{9}{16}\times\frac{1}{4}=\frac{9}{64}$

The amount of air remaining after removing the air third time $=\frac{9}{16}-\frac{9}{64}=\frac{9\times4-9}{64}=\frac{27}{64}$

The amount of air present in the cylinder when a vacuum pump removes each time $\frac{1}{4}$ of the remaining in the cylinder forms the below sequence

$1, \frac{3}{4}, \frac{9}{16}, \frac{27}{64}, ........$

$\frac{3}{4}-1=\frac{-1}{4}$, $\frac{9}{16}-\frac{3}{4}=\frac{-3}{16}$

$\frac{-1}{4}≠\frac{-3}{16}$

Therefore, the sequence of the numbers does not form an A.P.

(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.

The cost of digging the well for the first metre$= Rs.\ 150$.

The cost of digging the well for the second metre$= Rs.\ (150+50)=Rs.\ 200$.

The cost of digging the well for the third metre$= Rs.\ (200+50)=Rs.\ 250$.

The cost of digging the well for the fourth metre$= Rs.\ (250+50)=Rs.\ 300$.

The cost of digging the well for each successive metre (in rupees) forms the below sequence

$150, 200, 250, 300, ........$

Here, $a=150$ and $d=200-150=50$.

Therefore, the sequence of the numbers formed is an A.P.

(iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8% per annum.

Let

$a_1 = Rs.\ 10000$

This implies,

$a_2 = Rs.\ 10000 + Rs.\ 10000 \times \frac{8}{100}$

$= Rs.\ 10000 + Rs.\ 800$

$= Rs.\ 10800$

$a_3 = Rs.\ 10800 + Rs.\ 10800 \times 8100$

$= Rs.\ 10800 + Rs.\ 864$

$= Rs.\ 11664$

$a_2 - a_1 = Rs.\ 10800 - Rs.\ 10000$

$= Rs.\ 800$

$a_3 - a_2 = Rs.\ 11664 - ₹ 10800$

$= Rs.\ 864$

$a_3 - a_2 ≠ a_2 - a_1$

Therefore, the given information does not form an A.P.

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