# NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1

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.

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