# Arithmetic Progressions - NCERT Class 10 Maths

Arithmetic Progressions

Exercise 5.1

## Question 1

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.

### Solution

**Video 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.

## Question 2

Write first four terms of AP, When the first term \(a\) and the common difference \(d\) are given as follows:

(i) \(\begin{align} a = 10,\,d = 10\end{align}\)

(ii) \(\begin{align}a= - 2\quad , d = 0\end{align}\)

(iii) \(\begin{align}a= 4,\, d= - 3\end{align}\)

(iv) \(\begin{align}a = - 1,\,d = \frac{1}{2}\end{align}\)

(v) \(\begin{align}a = - 1.25,\,d = - 0.25\end{align}\)

### Solution

**Video Solution**

**Reasoning:**

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 difference.

**(i)** \(\begin{align} a = 10,\,d = 10\end{align}\)

**What is Known?**

\(\begin{align} a= 10\,\, {\rm{and}}\,\, d = 10\end{align}\)

**What is Unknown?**

First four terms of the AP.

**Steps:**

First term \(\begin{align}{a = 10}\end{align}\)

Second term \(\begin{align}{a+d = 10 + 10 = 20}\end{align}\)

Third term \(\begin{align}{a + 2d = 10 + 20 = 30}\end{align}\)

Fourth term \(\begin{align}{a + 3d = 10 + 30 = 40}\end{align}\)

The first four terms of AP are \(10, 20, 30,\) and \(40.\)

**(ii)** \(\begin{align}a= - 2,\quad d = 0\end{align}\)

**What is Known?**

\(\begin{align}a= - 2\,\, {\rm{and}}\,\, d = 0\end{align}\)

**What is Unknown?**

First four terms of the AP.

**Steps:**

First term \(\begin{align}{a =} -2\end{align}\)

Second term \(\begin{align}= a+ {d} = -2+0 = -2 \end{align}\)

Third term \(\begin{align}= a + {2d} = -2 +0 = -2 \end{align}\)

Fourth term \(\begin{align}= a + {3d} = -2 + 0 = -2 \end{align}\)

The first four terms of AP are \(-2,-2,-2\) and \(-2.\)

** (iii)** \(\begin{align}a= 4,\, d= - 3\end{align}\)

**What is Known?**

\(\begin{align} {a = 4}\,\, {\rm{and}}\,\, {d =} -3 \end{align}\)

**What is Unknown?**

First four terms of the AP.

**Steps:**

First term \(\begin{align} {= a = 4} \end{align}\)

Second term \(\begin{align} { a+d = 4+ (-3) = 1} \end{align}\)

Third term \(\begin{align} { a+2d} = 4 – 6= -2\end{align}\)

Fourth term \(\begin{align} { a+3d} = 4\; – 9 = -5\end{align}\)

The first four terms of AP are \(\begin{align} 4, 1, -2, -5\end{align}\).

**(iv) **\(\begin{align}a = - 1,\,d = \frac{1}{2}\end{align}\)

**What is Known?**

\(\begin{align}a = -1\,\, {\rm{and}}\,\, {d} =\frac{1}{2}\end{align}\)

**What is Unknown?**

First four terms of the AP.

**Steps:**

First term = \(a = - 1\)

Second term\(\begin{align}= {{a}} + {{d}} = - 1 + \frac{1}{2} = - \frac{1}{2}\end{align}\)

Third term \(\begin{align}= {{a}} + 2{{d}} = - 1 + 1 = 0\end{align}\)

Fourth term \(\begin{align}= {{a}} + 3{{d}} = - 1 + \frac{3}{2} = \frac{1}{2}\end{align}\)

The first four terms of AP are \(\begin{align} - 1, - \frac{1}{2},0,\frac{1}{2}\end{align}\) .

**(v)** \(\begin{align}a = - 1.25,\,d = - 0.25\end{align}\)

**What is Known?**

\(\begin{align}{a}= -1.25\,\, {\rm{and}}\,\, \quad {d} = - 0.25\end{align}\)

**What is Unknown?**

First four terms of the AP.

**Steps:**

First term\(\begin{align} = a = - 1.25\end{align}\)

Second term

\(\begin{align}&= {{a}} + {{d}}\\ &= - 1.25 + ( - 0.25)\\ &= - 1.25 - 0.25\\ &= - 1.5\\ \end{align}\)

Third term

\(\begin{align}&= a + 2d\\ &= - 1.25 + 2 \times ( - 0.25)\\ &= - 1.25 - 0.50\\ &= - 1.75\end{align}\)

Fourth term

\(\begin{align}&= a + 3{{d}}\\ &= - 1.25 + 3 \times ( - 0.25)\\ &= - 1.25 - 0.75\\ &= - 2.00\end{align}\)

The first four terms of AP are

\(\begin{align}-1.25, -1.5, -1.75, \rm{and} - 2.00\end{align}\).

## Question 3

For the following APs, write the first term and the common difference:

i) \(\begin{align}3,1, - 1, - 3 \ldots \ldots \end{align}\)

ii) \(\begin{align}- 5, - 1,3,7 \ldots \ldots \end{align}\)

iii) \(\begin{align}\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{{13}}{3} \ldots \ldots.\quad \quad \end{align}\)

iv) \(\begin{align}0.6,1.7,2.8,3.9 \ldots \end{align}\)

### Solution

**Video Solution**

**Reasoning:**

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 difference.

**i) What is known?**

\(3,1,-1,-3\) are in AP.

**What is Unknown?**

First term and the common difference of the AP.

**Steps:**

The AP is \(\begin{align} 3, 1 ,-1, -3.\end{align}\)

First term \(\begin{align}{a = 3}.\end{align}\)

Common difference

\[\begin{align} & = {a_2} - {a_1}\\&= 1 – 3\\&= -2\end{align}\]

First term is \(3\) and the common difference is \(-2.\)

**ii) What is known?**

\(\begin{align}-5, -1, 3, 7\end{align}\) are in AP

**What is Unknown?**

First term and the common difference of the AP.

**Step:**

The AP is \(\begin{align}-5, -1, 3, 7.\end{align}\)

First term\(\begin{align} a = - 5\end{align}\)

Common difference

\[\begin{align} & = {a_2} - {a_1}\\&= -1 – (-5)\\&= -1 + 5\\& = 4\end{align}\]

First term \(-5\) and the common difference is \(4.\)

**iii) What is known?**

\(\begin{align}\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{{13}}{3}\end{align}\) are in AP.

**What is Unknown?**

First term and the common difference of the AP.

**Steps:**

The AP is \(\begin{align}\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{{13}}{3}\end{align}\)

First term \(\begin{align} a = \frac{1}{3}\end{align}\)

Common difference

\[\begin{align} &= {a_2} - {a_1}\\&= \frac{5}{3} - \frac{1}{3}\\&= \frac{{5 - 1}}{3}\\&= \frac{4}{3}\end{align}\]

First term is \(\begin{align}\frac{1}{3}\end{align}\) and the common difference is \(\begin{align}\frac{4}{3}\end{align}\)

**iv) What is known?**

\(\begin{align}{\rm{0}}{\rm{.6,}}\,\,{\rm{1}}{\rm{.7,}}\,\,{\rm{2}}{\rm{.8,}}\,\,{\rm{3}} {\rm{.9}}\end{align}\) are in AP.

**What is Unknown?**

First term and the common difference of the AP.

**Steps:**

The AP is \(\begin{align} {\rm{0}}{\rm{.6,}}\,\,{\rm{1}}{\rm{.7,}}\,\,{\rm{2}}{\rm{.8,}}\,\,{\rm{3}}{\rm{.9}}\end{align}\)

First term \(\begin{align} a = 0.6 \end{align}\)

Common difference

\[\begin{align} &= {a_2} - {a_1}\\&= 1.7 – 0.6\\&= 1.1\end{align}\]

First term \(= 0.6\) and the common difference \(= 1.1\)

## Question 4

Which of the followings are APs? If they form an AP, Find the common difference d and write three more terms.

i) \(\begin{align}2,4,16\dots\dots\end{align}\)

ii)\(\begin{align}2,\frac{5}{2},3,\frac{7}{2}\dots\dots\end{align}\)

iii) \(\begin{align}- 1.2, - 3.2, - 5.2, - 7.2\dots\dots\end{align}\)

iv)\(\begin{align}{\kern 1pt} \,\,\, - 10, - 6, - 2,2\dots\dots\end{align}\)

v)\(\begin{align}3,3 + \sqrt 2 ,\,3 + 2\sqrt 2 ,3 + 3\sqrt 2 \dots\dots\end{align}\)

vi) \(\begin{align}0.2,0.22,0.222,0.2222\dots\dots\end{align}\)

vii)\(\begin{align}0, - 4, - 8, - 12\dots\dots\end{align}\)

viii)\(\begin{align}- \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}\dots\dots\end{align}\)

ix) \(\begin{align}1,3,9,27\dots\dots\end{align}\)

x)\(\begin{align}a,2a,3a.4a,\dots\dots\end{align}\)

xi) \(\begin{align}a,{a^2},{a^3},{a^4}\end{align}\)

xii) \(\begin{align}\sqrt 2 ,\sqrt 8 ,\sqrt {18} ,\sqrt {132} \end{align}\)

xiii)\(\begin{align}\sqrt 3 ,\sqrt 6 ,\sqrt 9 ,\sqrt {12} \end{align}\)

xiv)\(\begin{align}{1^2},{3^2},{5^2},{7^2}\end{align}\)

xv)\(\begin{align}{1^2},{3^2},{5^2},{7^3}\end{align}\)

### Solution

**Video Solution**

** Reasoning:**

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.

**i) What is Known?**

\(\begin{align} 2, 4, 8, 16…..\end{align}\)

**What is Unknown?**

Common difference and next three more terms of AP if it is an AP.

**Steps:**

The given numbers are \(2,\,4,\,8,\,16\)

First term \( a =2\)

Common difference

\(d=a_{2}-a_{1}=4-2=2\)

Common difference

\(d=a_{3}-a_{2}=8-4=4\)

\(\begin{align}({a_{3}-a_{2}) \neq (a_{2}-a_{1})}\end{align}\)

\(\begin{align}2, 4, 8, 16\end{align}\) are not in AP, because the common difference is not equal.

**ii) What is Known?**

\(\begin{align}2,\,\frac{5}{2},\,3,\,\frac{7}{2}\end{align}\)

**What is Unknown?**

Common difference and next three more terms of AP if it is an AP.

**Steps:**

The given numbers are \(\begin{align}2,\,\frac{5}{2},\,3,\,\frac{7}{2}\end{align}\)

First term a \(= 2\)

Common difference

\[\begin{align}& d =a_{2}-a_{1}=\frac{5}{2}-2=\frac{5-4}{2}=\frac{1}{2}\end{align}\]

Common difference

\[\begin{align}& d =a_{3}-a_{2}=3-\frac{5}{2}=\frac{6-5}{2}=\frac{1}{2}\end{align}\]

Since

\(\begin{align}({a_3} - {a_2}) = ({a_2} - {a_1}).\end{align}\)

\(\begin{align}2,\,\frac{5}{2},\,3,\,\frac{7}{2}\end{align}\) forms an AP and common difference is\(\begin{align}\frac{1}{2}\end{align}\) .

The next three terms are:

Fifth term

\[\begin{align} &=a+4 d \\ &=2+4 \times \frac{1}{2} \\ &=2+2 \\ &=4 \end{align}\]

Sixth term

\[\begin{align} &=a+5 d \\ &=2+5 \times \frac{1}{2} \\ &=2+\frac{5}{2} \\ &=\frac{4+5}{2} \\ &=\frac{9}{2} \end{align}\]

Seventh term

\[\begin{align} &=a+6 d \\ &=2+6 \times \frac{1}{2} \\ &=5 \end{align}\]

\(\begin{align}2,\,\frac{5}{2},\,3,\,\frac{7}{2}\end{align}\) forms an AP and the common difference is \(\begin{align}\frac{1}{2}\end{align}\). The next three terms are \(\begin{align}4,\,\frac{9}{2}\,,5.\end{align}\)

**iii) What is Known?**

\(\begin{align} - 1.2,\, - 3.2,\, - 5.2,\, - 7.2 \ldots \ldots .\end{align}\)

**What is Unknown?**

Whether it forms an AP. If it is find the common difference and the next three terms of AP.

**Steps:**

The given numbers are

\(\begin{align} - 1.2,\, - 3.2,\, - 5.2,\, - 7.2 \ldots \ldots .\end{align}\)

First term \(\begin{align} a = -1.2\end{align}\)

Common difference

\[\begin{align} d &={a_2} - {a_1}\\& = - 3.2 - ( - 1.2)\\& = - 3.2 + 1.2 = - 2\end{align}\]

Common difference

\[\begin{align} &={a_3} - {a_2} = - 5.2 - ( - 3.2)\\ &= - 5.2 + 3.2 = - 2 \end{align}\]

Since

\(\begin{align}({a_3} - {a_2}) = ({a_2} - {a_1}) . \end{align}\)

It forms an AP.

The fifth term

\[\begin{align} &=a+4d \\ &=-1.2+4(-2) \\ &=-1.2-8=-9.2 \end{align}\]

The sixth term

\[\begin{align} &=a+5d \\ &=-1.2+5(-2) \\ &=-1.2-10 \\ &=-11.2 \end{align}\]

The seventh term

\[\begin{align} &=a+6d \\ &=-1.2+6(-2) \\ &=-1.2-12 \\ &=-13.2 \end{align}\]

\(\begin{align} - 1.2,\, - 3.2,\, - 5.2,\, - 7.2\end{align}\) forms an AP with common difference -2. The next three terms of AP are \(\begin{align}-9.2, -11.2, -13.2. \end{align}\)

**iv) What is Known?**

\(\begin{align} -10, -6, -2, 2 \end{align}\)

**What is Unknown?**

Whether it forms an AP. If it is find the common difference and the next three terms of AP.

**Steps:**

The given numbers are \(\begin{align} -10, -6, -2, 2 \end{align}\)

First term a \(= -10\)

Common difference

\[\begin{align} d &=a_{2}-a_{1} \\ &=-6-(-10) \\ &=-6-10 \\ &=4 \end{align}\]

Common difference

\[\begin{align} d &=a_{3}-a_{2} \\ &=-2-(-6) \\ &=-2+6 \\ &=4 \end{align}\]

Since

\((a_{3}-a_{2})=(a_{2}-a_{1})\)

Fifth Term: \( a+4 d=-10+16=6\)

Sixth Term: \(a+5 d=-10+20=10\)

Seventh Term: \(a+6 d=-10+24=14\)

\(\begin{align}-10, -6 ,-2, 2 \end{align}\) forms an AP with common difference \(4\) and next terms are \(6, 10, 14.\)

**v) What is Known?**

\(\begin{align}3,\,\,3 + \sqrt 2 \,\,,3 + 2\sqrt 2 \,\,,3 + 3\sqrt 2 \ldots \ldots \end{align}\)

**What is Unknown?**

Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.

**Steps:**

The list of numbers is

\(\begin{align}3,\,\,3 + \sqrt 2 \,\,,3 + 2\sqrt 2 \,\,,3 + 3\sqrt 2 \ldots \ldots \end{align}\)

Common difference

\[\begin{align} d &=a_{2}-a_{1} \\ &=3+\sqrt{2}-3 \\ &=\sqrt{2} \end{align}\]

Common difference

\[\begin{align} d &=a_{3}-a_{2} \\ &=3+2 \sqrt{2}-(3+\sqrt{2}) \\ &=3+2 \sqrt{2}-3-\sqrt{2} \\ &=\sqrt{2} \end{align}\]

Since

\(\begin{align} a_{3}-a_{2}=a_{2}-a_{1}\end{align}\)

So \(\begin{align}3,\,\,3 + \sqrt 2 \,\,,3 + 2\sqrt 2 \,\,,3 + 3\sqrt 2 \ldots \ldots \end{align}\) forms an AP with common difference 4.

Next three terms are

\(\begin{align} \text { Fifth term } &=a+4 d \\ &=3+4 \sqrt{2} \\ \text { Sixth term } &=a+5 d \\ &=3+5 \sqrt{2} \\ \text { Seventh term } &=a+6 d \\ &=3+6 \sqrt{2} \end{align}\)

It is an AP with common difference \(\begin{align}\sqrt 2 \end{align}\) and Next three terms are

\(\begin{align}3 + 4\sqrt 2 ,\,\,3 + 5\sqrt 2 \,\,,3 + 6\sqrt 2 \end{align}\) ,.

**vi) What is Known?**

\(\begin{align}0.2,0.22,0.222,0.2222.....\end{align}\)

**What is Unknown?**

Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.

**Steps:**

The list of numbers are

\(\begin{align}0.2,0.22,0.222,0.2222.....\end{align}\)

Common difference

\[\begin{align} d &=a_{2}-a_{1} \\ &=0.22-0.2 \\ &=0.20 \end{align}\]

Common difference

\[\begin{align} d &=a_{3}-a_{2} \\ &=0.222-0.220 \\ &=0.002\\ (a_{3}-a_{2} )&\neq (a_{2}-a_{1})\end{align}\]

The given list of numbers does not form an AP.

**vii) What is Known?**

\(\begin{align} 0,-4,-8,-12……\end{align}\)

**What is Unknown?**

Given list of numbers form an AP or not. If it is find the common difference and the next three terms of AP.

**Steps:**

The list of numbers is

\(\begin{align} 0,-4,-8,-12……\end{align}\)

Common difference

\(\begin{align}d = {a_2} - {a_1}= - 4 – 0 = -4\end{align}\)

Common difference

\(\begin{align}d = {a_3} - {a_2}= -8 – (-4) = -8 + 4 = -4\end{align}\)

Since\(\begin{align}({a_3} - {a_2}) = ({a_2} - {a_1})\end{align}\) . It forms an AP.

\(\begin{align} \text { Fifth term } &=a+4 d \\ &=0+4(-4) \\ &=-16 \\ \text { Sixth term } &=a+5 d \\ &=0+5(-4) \\ &=-20 \\ \text { Seventh term } &=a+6 d \\ &=0+6(-4) \\ &=-24 \end{align}\)

The given numbers form an AP with difference \(-4.\) The next three terms are \(\begin{align} -16, -20, -24. \end{align}\)

**viii) What is Known?**

\(\begin{align} - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}.......\end{align}\)

**What is Unknown?**

**Steps:**

The given list of numbers is

\(\begin{align} - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}.......\end{align}\)

Common difference

\[\begin{align} d &= {a_2} - {a_1}\\&= - \frac{1}{2} - \left( { - \frac{1}{2}} \right)\\ &= - \frac{1}{2} + \frac{1}{2}\\&= 0\end{align}\]

Common difference

\[\begin{align} d &= {a_3} - {a_2}\\& = - \frac{1}{2} - \left( { - \frac{1}{2}} \right)\\ &= - \frac{1}{2} + \frac{1}{2}\\&= 0\end{align}\]

Since \(\begin{align}({a_3} - {a_2}) = ({a_2} - {a_1})\end{align}\) .The list of numbers forms an AP.

\(\begin{align}{\text{The fifth term }} &= a + 4d\\& = - \frac{1}{2} + 4(0)\\ &= - \frac{1}{2}\\{\text{ The sixth term }} &= a + 5d\\ &= - \frac{1}{2} + 5(0)\\ &= - \frac{1}{2}\\{\text{The seventh term}} &= {a} + 6d\\ &= - \frac{1}{2} + 6(0)\\ &= - \frac{1}{2}\end{align}\)

The given list of numbers form an AP with common difference \(d = 0.\) Next three terms are \(\begin{align} - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}.\end{align}\)

**ix) What is Known?**

\(\begin{align}1, 3 ,9 ,27.\end{align}\)

**What is Unknown?**

**Steps:**

The list of numbers are \(\begin{align}1, 3 ,9 ,27.\end{align}\)

Common difference \(\begin{align} d = {a_2} - {a_1}= 3 – 1 = 2\end{align}\)

Common difference \(\begin{align} d = {a_3} - {a_2}= 9 -3 =6\end{align}\)

Since\(\begin{align} ({a_2} - {a_1}) \ne ({a_3} - {a_2})\end{align}\)

The given list of numbers does not form an AP.

**x) What is Known?**

\(\begin{align}{a, 2a, 3a, 4a}\end{align}\)

**What is Unknown?**

**Steps:**

The list of numbers is \(\begin{align}{a, 2a, 3a, 4a}.\end{align}\)

Common difference

\[\begin{align} d &=a_{2}-a_{1} \\ &=2a-a\\ &=a \end{align}\]

Common difference

\[\begin{align} d &=a_{3}-a_{2} \\ &=3a-2a \\ &=a \end{align}\]

since\(\begin{align}{a_3} - {a_2} = {a_2} - {a_1}, {a, 2a, 3a, 4a}\end{align}\)

forms an AP.

The fifth term \(=a+4d=a+4a=5a\)

The sixth term \(=a+5d=a+5a=6a\)

The seventh term \(=a+6d=a+6a=7a\)

The given list of numbers form an AP with common difference \(d =a\) The next three terms are \(5a, 6a, 7a.\)

**xi) What is Known?**

\(\begin{align}a,{a^2},{a^3},{a^4}\end{align}\)

**What is Unknown?**

**Steps:**

The list of numbers is \(\begin{align}a,{a^2},{a^3},{a^4}\end{align}\)

Common difference

\[\begin{align} d &=a_{2}-a_{1} \\ &=a^{2}-a \\ &=a(a-1) \end{align}\]

Common difference

\[\begin{align} d &= {a_3} - {a_2}\\ &= {a^3} - {a^2}\\ &= {a^2}\left( {a - 1} \right)\end{align}\]

Since

\(\begin{align}({a_2} - {a_1}) \ne ({a_3} - {a_2})\end{align}\)

The given list of numbers does not form an AP.

**xii) What is Known?**

\(\begin{align}\sqrt 2 ,\sqrt 8 ,\sqrt {18} ,\sqrt {32} \end{align}\)

**What is Unknown?**

**Steps:**

The list of numbers is \(\begin{align}\sqrt 2 ,\sqrt 8 ,\sqrt {18} ,\sqrt {32} \end{align}\)

Common difference

\[\begin{align}d &={a_2} - {a_1}\\&= \sqrt 8 - \sqrt 2 \\ &= \sqrt {4 \times 2} - \sqrt 2 \\ &= 2\sqrt 2 - \sqrt 2 \\ &= \sqrt 2 \end{align}\]

Common difference

\[\begin{align}d &= {a_3} - {a_2}\\ &= \sqrt {18} - \sqrt 8 \\ &= \sqrt {9 \times 2} - \sqrt {4 \times 2} \\ &= 3\sqrt 2 - 2\sqrt 2 \\& = \sqrt 2 \end{align}\]

since

\(\begin{align}({a_2} - {a_1}) = ({a_3} - {a_2})\end{align}\) The given numbers form an AP.

\(\begin{align}{\text{The fifth term }} &= a + 4d\\ &= \sqrt 2 + 4\sqrt 2 \\ &= 5\sqrt 2 \\ &= \sqrt {25 \times 2} \\ &= \sqrt {50} \\{\text{The sixth term}}& = a + 5d\\ &= \sqrt 2 + 5\sqrt 2 \\ &= 6\sqrt 2 \\ &= \sqrt {36 \times 2} \\ &= \sqrt {72} \\{\text{ The seventh term }} &= {\rm{a}} + 6d\\ &= \sqrt 2 + 6\sqrt 2 \\ &= 7\sqrt 2 \\ &= \sqrt {49 \times 2} \\ &= \sqrt {98} \end{align}\)

The list of numbers forms an AP with common difference\(\begin{align}\sqrt 2 \end{align}\). Next three terms are \(\begin{align}\sqrt {50} ,\sqrt {72} ,\sqrt {98} \end{align}\)

**xiii ) What is Known?**

\(\begin{align}\sqrt 3 ,\,\sqrt 6 \,,\sqrt 9 ,\,\sqrt {12} \end{align}\)

**What is Unknown?**

**Steps:**

The list of numbers is

\(\begin{align}\sqrt 3 ,\,\sqrt 6 \,,\sqrt 9 ,\,\sqrt {12} \end{align}\)

Common difference

\[\begin{align}d &= {a_2} - {a_1} = \sqrt 6 - \sqrt 3 \\ &= \sqrt {3 \times 2} - \sqrt 3 \\ &= \sqrt 3 \left( {\sqrt 2 - 1} \right)\end{align}\]

Common difference

\[\begin{align} d&= {a_3} - {a_2} = \sqrt 9 - \sqrt 6 \\ &= \sqrt {3 \times 3} - \sqrt {3 \times 2} \\ &= \sqrt 3 (\sqrt 3 - \sqrt 2 )\end{align}\]

since\(({a_2} - {a_1}) \ne ({a_3} - {a_2})\)

The given list of numbers does not form an AP.

**xiv) What is Known?**

\(\begin{align}{1^2},\,{3^2}\,,{5^2}\,,{7^2}\end{align}\)

**What is Unknown?**

**Steps:**

The list of numbers is

\(\begin{align}{1^2},\,{3^2}\,,{5^2}\,,{7^2}\end{align}\)

Common difference

\(\begin{align} d= {a_2} - {a_1}= 25 – 1 = 24\end{align}\)

Common difference

\( d ={a_3} - {a_2}= 49 – 25 = 24\)

Since \(({a_2}- {a_1}) = ({a_3} - {a_2})\) they form an AP

\(\begin{align} \text { The fifth term } &=a+4 d \\ &=1+4 \times 24 \\ &=1+96 \\ &=97 \end{align}\)

\(\begin{align} \text { The sixth term } &=a+5 d \\ &=1+5 \times 24 \\ &=1+120 \\ &=121 \end{align}\)

\(\begin{align} \text { The seventh term } &=a+6 d \\ &=1+6 \times 24 \\ &=1+144 \\ &=145 \end{align}\)

The list of numbers form an AP with common difference \(24.\) The next three terms are \(97, 121,\) and \(145.\)

**xv)What is Known?**

\(\begin{align}{1^2},\,{3^2},\,{5^2},\;{7^3}\end{align}\)

**What is Unknown?**

**Steps:**

The list of numbers

\(\begin{align}{1^2},\,{3^2},\,{5^2},\;{7^3}\end{align}\)

Common difference

\[\begin{align} d = {a_2} - {a_1}= 9 – 1 = 8\end{align}\]

Common difference

\[\begin{align}d = {a_3} - {a_2}= 25 - 9 = 16\end{align}\]

Since

\(\begin{align}({a_2} - {a_1}) \ne ({a_3} - {a_2})\end{align}\)

The given list of numbers does not form an AP.

In the introduction part of the chapter 5, the concept of Arithmetic Progressions is explained - patterns in which succeeding terms are obtained by adding a fixed number to the preceding terms. Thus, an arithmetic progression is defined as a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. Further, it is noted that the fixed number is called as Common Difference and it can be positive, negative or zero. Furthermore, the general form of an AP is discussed and infinite and finite forms of AP are explained. Thereafter, for the second exercise of the chapter, the formula for finding the nth term of an AP is derived. Finally, the formula for finding the sum of the first n terms of an AP is derived.

**Download Chapter 5 Arithmetic Progressions Class 10 Maths Solutions PDF for FREE:**

## Exercise 5.1

## Exercise 5.2

## Exercise 5.3

## Exercise 5.4

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