Question

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

Answer 1

Hint:A sequence of numbers is called an Arithmetic progression if the difference between any two consecutive terms is always the same.Check the difference between the consecutive terms of the given sequence. If the difference between any two consecutive terms is not the same, then it is not an AP.

Complete step-by-step answer:
An arithmetic progression is given by $a,\left( {a + d} \right),\left( {a + 2d} \right),\left( {a + 3d} \right),......$
Where $a$ is the first term of AP and $d$ is the common difference of AP.
Let us analyse the four situations.
(1) Given, Fare of first km$ = Rs.15$
Fare for each additional km$ = Rs.8$
We can write the given conditions as,
 Taxi fare for 1st km, ${a_1} = 15$
Taxi fare for first 2 km, ${a_2} = 15 + 8 = 23$
Taxi fare for first 3 km, ${a_3} = 23 + 8 = 31$
Taxi fare for first 4 km, ${a_4} = 31 + 8 = 39$
And so on…
The list of numbers is $15,23,31,39,....$
Here, ${a_2} - {a_1} = 23 - 15 = 8$
${a_3} - {a_2} = 31 - 23 = 8$
${a_4} - {a_3} = 39 - 31 = 8$
The difference between the successive terms are the same i.e., $8$. Hence, the above list of numbers forms an AP.

(2) Let the initial volume of air in a cylinder be $V$ liters. In each stroke, the vacuum pump removes $\dfrac{1}{4}$ of the air remaining in the cylinder at a time. In other words, after every stroke, only $1 - \dfrac{1}{4} = \dfrac{3}{4}th$ part of air will remain.
Hence the series can be written as below:
$V,\dfrac{{3V}}{4},\dfrac{{3V}}{4} - \dfrac{1}{4}\left( {\dfrac{{3V}}{4}} \right),....$
$ = V,\dfrac{{3V}}{4},\dfrac{{12V - 3V}}{{16}},....$
$ = V,\dfrac{{3V}}{4},\dfrac{{9V}}{{16}},....$
Now, common difference,
\[d = Second{\text{ }}term - first{\text{ }}term = \dfrac{{3V}}{4} - V = \dfrac{{ - V}}{4}\]
$d = Third{\text{ }}term - Second{\text{ }}term = \dfrac{{9V}}{{16}} - \dfrac{{3V}}{4} = \dfrac{{9V - 12V}}{{16}} = \dfrac{{ - 3V}}{{16}}$
From the above values of $d$, the common difference is not the same between the series. Hence, this is not an AP.

(3) Given, cost of digging for first meter $ = Rs.150$
Rise in cost for each subsequent meter$ = Rs.50$
We can write the given conditions as,
 Cost of digging for 1st meter, ${a_1} = 150$
Cost of digging for first 2 meters, ${a_2} = 150 + 50 = 200$
Cost of digging for first 3 meters, ${a_3} = 200 + 50 = 250$
Cost of digging for first 4 meters, ${a_4} = 250 + 50 = 300$
And so on…
The list of numbers is $150,200,250,300,....$
Here, ${a_2} - {a_1} = 200 - 150 = 50$
${a_3} - {a_2} = 250 - 200 = 50$
${a_4} - {a_3} = 300 - 250 = 50$
The difference between the successive terms are the same i.e., $50$. Hence, the above list of numbers forms an AP.

(4) We know that if Rs. $P$ is deposited at $r\% $ compound interest per annum for $n$ years, our money will be $P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$ after $n$ years.
Given: rate, $r = 8\% $ per annum and principal, $P = $Rs. $10000$
Therefore, after every year, the money will be
$10000,10000{\left( {1 + \dfrac{8}{{100}}} \right)^1},10000{\left( {1 + \dfrac{8}{{100}}} \right)^2},10000{\left( {1 + \dfrac{8}{{100}}} \right)^3},....$
$10000,10000\left( {\dfrac{{108}}{{100}}} \right),10000{\left( {\dfrac{{108}}{{100}}} \right)^2},10000{\left( {\dfrac{{108}}{{100}}} \right)^3},....$
$10000,10800,11664,....$
Now, common difference,
\[d = Second{\text{ }}term - first{\text{ }}term = 10800 - 10000 = 800\]
$d = Third{\text{ }}term - Second{\text{ }}term = 11664 - 10800 = 864$
From the above values of $d$, the common difference is not the same between the series. Hence, this is not an AP.

So in the above given statements $(i)$ and $(iii)$ forms arithmetic progression as the difference between the successive terms are the same.

Note:In an AP, every succeeding term is obtained by adding $d$ to the preceding term. So, $d$ can be found by subtracting any term from its succeeding term. To check whether given terms are in AP or not, we need to find at least any two of${a_2} - {a_1},{a_3} - {a_2},{a_4} - {a_3},....$. It is not enough to find only one of them.