# Ex.5.1 Q1 Arithmetic progressions Solutions - NCERT Maths Class 10

## 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 \(\rm{}km,\) when the fare is \( \rm{Rs.} \,15\) for the first km and \(\rm{Rs.} \,8\) for each of additional \(\rm{}km.\)

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

(iii) The cost of digging a well after every meter of digging, when it costs \(\rm{Rs}\,150\) for the first meter and rises by \(\rm{Rs.}\, 50\) for each subsequent meter.

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

## Text Solution

(i)

**What is Known?**

Charges for the first km and additional \(\rm{km.}\)

**What is Unknown?**

Whether it is an arithmetic progression or not ?

**Reasoning:**

An arithmetic progression is a sequence of number in which each term is obtained by adding a fixed number to the preceding term except the first term.

General form of an arithmetic progression is \(\begin{align}a, \,(a+d),\,( a+2d),\, (a+3d),\end{align}\) Where

\(a\) is the first term and \(d\) is the common difference.

**Steps:**

Taxi fare for \(1 \,\rm{km}\) \(\begin{align}= {\rm{Rs.}}\,15\left( {{a_1}} \right)\end{align}\)

Taxi fare for \(2 \,\rm{km}\) \(\begin{align}= 15+8 = {\rm{Rs}}. 23\left( {{a_2}} \right)\end{align}\)

Taxi fare for \(3\, \rm{km}\)

\(\begin{align}= 15+8+8 = \,{\rm{Rs.}} \,31\left( {{a_3}} \right)\end{align}\)

And so on.

\[\begin{align}\left( {{a_2}} \right) - \left( {{a_1}} \right) &= \rm{Rs}(23 -15) = \rm{Rs.} \,8\\\left( {{a_3}} \right) - \left( {{a_2}} \right)&= \rm{Rs}(31 - 23) = \rm{Rs. 8}\end{align}\]

Every time the difference is same.

So, this forms an AP with first term \(15\) and the difference is \(8.\)

(ii)

**What is Known?**

The amount of air present.

**What is Unknown?**

Whether it is an arithmetic progression or not?

**Reasoning:**

An arithmetic progression is a sequence of number in which each term is obtained by adding a fixed number to the preceding term except the first term.

General form of an arithmetic progression is \(\begin{align}a, \,(a+d),\,( a+2d),\, (a+3d),\end{align}\) Where

\(a\) is the first term and \(d\) is the common difference.

**Steps:**

Let the amount of air in the cylinder be \(x.\)

So \(\begin{align}{a_1= x}\end{align}\)

After First time removal,

\[\begin{align}{a_2}&= x - \frac{x}{4}\\{a_2}&= \frac{{3x}}{4}\end{align}\]

After Second time removal,

\[\begin{align}{\text{}}\, {a_3}&= \frac{3}{4} - \frac{1}{4}\left( {\frac{{3x}}{4}} \right)\\&= \frac{{3x}}{4} - \frac{{3x}}{{16}}\\&= \frac{{12x - 3x}}{{16}}\\&= \frac{{9x}}{{16}}\end{align}\]

\[\begin{align}{a_3} = \frac{{9x}}{{16}}\end{align}\]

After third time removal,

\[\begin{align}a_4&= \frac{{9x}}{{16}} - \frac{1}{4}\left( {\frac{{9x}}{{16}}} \right)\\ &= \frac{{9x}}{{16}} - \frac{{9x}}{{64}} \\&= \frac{{36x - 9x}}{{64}} \\&= \frac{{27x}}{{64}}\end{align}\]

\[\begin{align}{a_4} &= \frac{27x}{64}\\{a_2} - {a_1} &= \frac{{3x}}{4} - x \\&= \frac{{3x - 4x}}{4} \\&= \frac{{ - x}}{4}\\{a_3} - {a_2} &= \frac{9}{{16}}x - \frac{3}{4}x \\&= \frac{{9x - 12x}}{{16}} \\&= \frac{{ - 3x}}{{16}}\\({a_3} - {a_2}) &\ne ({a_2} - {a_1})\end{align}\]

This is not forming an AP.

iii)

**What is Known?**

Cost of digging well for every meter and subsequent meter.

**What is unknown?**

Whether it is an arithmetic progression or not.

**Reasoning:**

An arithmetic progression is a sequence of number in which each term is obtained by adding a fixed number to the preceding term except the first term.

General form of an arithmetic progression is \(\begin{align}a, \,(a+d),\,( a+2d),\, (a+3d),\end{align}\) Where

\(a\) is the first term and \(d\) is the common difference.

**Steps:**

Cost of digging the well after \(1\) meter

\(\begin{align}= {\rm{Rs.}} \,150\left( {{a_1}} \right)\end{align}\)

Cost of digging the well after \(2\) meters

\(\begin{align}= 150 + 50 = {\rm{Rs.}}\, 200\,\left( {{a_2}} \right)\end{align}\)

Cost of digging the well after \(3\) meters

\(=\rm{ Rs.}\, 150+50+50={Rs.\,}250\) \(\begin{align}\left( {{a_3}} \right)\end{align}\)

\[\begin{align}\left( {{a_2} - {a_1}} \right) &= 200 - 150 = {\rm{Rs}}.50\\\left( {{a_3} - {a_2}} \right) &= 250 - 200 = {\rm{Rs}}.50\\\left( {{a_2} - {a_1}} \right) &= \left( {{a_3} - {a_2}} \right)\end{align}\]

So, this list of numbers from an AP with first term as \(\rm{Rs.} \,150\) and common difference is \(\rm{Rs.}\, 50.\)

iv)

**What is Known?**

Deposited amount and rating interest.

**What is Unknown?**

Whether it is an arithmetic progression or not.

**Reasoning:**

\(a\) is the first term and \(d\) is the common difference.

**Steps:**

Amount present when the amount is \(P\) and the interest is \(r \%\) after \(n\) years is

\[\begin{align}\mathrm{A}&=\left[P\left(1+\frac{r}{100}\right)\right] \\ \mathrm{P}&=10,000 \\ \mathrm{r}&=8 \%\end{align}\]

For first year \(\begin{align}( a_1) = 10000\, \left( 1 + \frac{8}{{100}}\right)\end{align}\)

For second year

\(\begin{align}\left( {{a_2}} \right) = 10000{{\left( {1 + \frac{8}{{100}}} \right)}^2}\\\end{align}\)

For third year \(\begin{align}\left( {{a_3}} \right) = 10000{{\left( {1 + \frac{8}{{100}}} \right)}^3}\end{align}\)

For fourth year \(\begin{align}\left( {{a_4}} \right) = 10000{{\left( {1 + \frac{8}{{100}}} \right)}^4}\end{align}\)

And so on

\(\begin{align}&\left( {{a_2}\! - \!{a_1}} \right) \\ \!&=\! 10000{\left( {1 \!+\! \frac{8}{{100}}} \right)^2}\! -\! 10000\left( {1 \!+\! \frac{8}{{100}}} \right)\\ &\!=\! 10000\left( {1 \!+\! \frac{8}{{100}}} \right)\left[ {1 \!+\! \frac{8}{{100}} \!-\! 1} \right]\\ &= 10000\left( {1 + \frac{8}{{100}}} \right)\left( {\frac{8}{{100}}} \right)\\ &\left( {{a_3} - {a_2}} \right) \\&=\! 10000{\left( {1 \!+\! \frac{8}{{100}}} \right)^3}\! \!-\! 10000{\left(\! {1 \!+ \!\frac{8}{{100}}} \!\right)^2}\\ &= 10000{\left( {1 + \frac{8}{{100}}} \right)^2}\left[ {1 + \frac{8}{{100}} - 1} \right]\\ &= 10000{\left( {1 + \frac{8}{{100}}} \right)^2}\left( {\frac{8}{{100}}} \right)\\&\left( {{a_3} - {a_2}} \right) \ne \left( {{a_2} - {a_1}} \right)\end{align}\)

The amount will not form an AP.