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

Go back to  'Ex.5.1'

## 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.

Video Solution
Arithmetic Progressions
Ex 5.1 | Question 1

## 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:

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:

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.

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school