Here, you will get a complete in-depth article for class 10 arithmetic progressions. You are going to learn about arithmetic progression class 10 notes, their ncert solution, and answer to the important question . For the rest of the topic, you can see in the table of content given below where topics are categorized according to arithmetic progression class 10 notes,ncert solution respectively . You can easily access that topic

## Arithmetic progression class 10 notes:[Progression]

**Progression: It is the collection of numbers that are arranged according to a definite rule.**

Here are examples:

1. 2,4,6,8,10…………. ( Rule. every successive term has a difference of 2 with the previous term)

2. 2,4,8,16,32…………. ( Rule. every successive term is 2 times of the previous term)

3. 3,5,7,9,11,13………… ( Rule. every successive term has a difference of 2 with the previous term)

4. 1/3,1/5, 1/7,1/9 ……….

In example** (4)** It seems to be there is not any definite rule but here is Rule: Denominator of every consecutive term has the difference of 2 (3,5,7,9……..)

### Arithmetic progression class 10:[Types of progressions]

There are three types of progression

**(i)** **Arithmetic progression(A.P)**

**(ii) Geometric progression(G.P)**

**(iii) Harmonic progression(H.P)**

we will study about G.P and H.P in higher classes. But I will cover its basic in this article so that you know a little bit about them.

**Arithmetic progression:** Progression in which the difference between consecutive terms is always constant is called an arithmetic progression.

example:(i) 2,4,6,8,……………… is arithmetic progression because the difference between every consecutive term is constant i.e 2.

General form of AP are a,a+d,a+2d,a+3d…..

Arithmetic progressions(A.P) are generally of two types.

(i) Finite A.P

(ii) Infinite A.P

**Finite A.P**= Those A.P’s which have a finite number of terms are called finite A.P.

For example.2,4,6,8,……42

**Infinite A.P**=Those A.P’s which do not have a finite number of terms is called Infinite A.P.

For example

2,4,6,8,10………..

**Geometric progression**: The progression in which the ratio of a term to the preceding term to it, is always constant

Examples. 2,4,8,16,32………..etc

Here \frac{second term}{ first term}=\frac{4}{2}=2

similarly \frac{third term}{ second term}=\frac{8}{4}=2 …….so on

**Harmonic progression:The progression which is formed through the reciprocal of the arithmetic progression **

In the above example, we have A.P as 2,4,6,8,………………

Reciprocal of above AP will be 1/2, 1/4,1/6,1/8 ………..

These types of progression known as Harmonic progression

Now, let’s come to our original topic ” **arithmetic progression class 10** ”

Important notation in **arithmetic progression class 10 **

**a= First term of A.P**

**d=common difference of A.P**

** a_{n}= nth term of A.P**

s_{n}=Sum of first n term of Arithmetic progression

l= Last term of an arithmetic progression

**Arithmetic progression class 10 ** problems

Q.1 Is the sequence 1,11,21,31,……an arithmetic progression?

Ans. Yes, it is arithmetic progression because every term has a difference of 10

Q.2 Is the sequence 2,4,6,10,14…..an arithmetic progression?

Ans. No, because every term do not have a difference of 2 (10-6=4,14-10=4)

Q.3 Is the sequence 1/3,1/5,1/7,1/9……..an A.P?

Ans. No, it is not A.P because the difference between terms is not constant. It is an example of Harmonic progression’.

### Arithmetic progression class 10 notes:[n-term of an A.P]

If a is the first term and d is a common difference then its n-term is given by the formula

a_{n}=a+( n-1) d

There is another concept of n-term in which we find a term from the end not from the beginning. Here is an example

a_{1} ,a_{2},a_{3}……………l is an AP

a_{1}=First term

l=last term

Let d_{1}=common difference

To find any term from the end, simply reverse the AP

l………..a_{3},a_{2},a_{1}

Here,

l=first term

-d_{1}=common difference (common difference become negative when we reverse any AP)

Its nth term will be given by,

a_{n}=l+(n-1)(-d_{1})

Let’s take some examples to understand it clearly

Q.1 Find the 10th term of the A.P. 1,4,7,10……

Sol. We have first term (a) = 1

Common difference(d)=3

n=10

subtitute these value in equation a_{n} =a+( n-1) d

a_{10}=1+(10-1)3

a_{10}=1+27

a_{10}=28

Q.2 Is 68 a term of the A.P 7 ,10,13,……….?

Sol. Let 68 is n-term of given A.P

a=7

d=3

a_{n} =68

Substitute given value in the equation

a_{n} =a+( n-1) d

68=7+(n-1) 3

68=7+3n-3

68=4+3n

64=3n

n=64/3

As you can see, here the value of n is fractional but n should be the natural number for A.P

so 68 is not the term of given A.P

Q.3 The first term of an A.P is 5, the common difference is 3 and the last term is 80; find the number of terms

sol. We have a=5 ,d=3 and l=80

As we know l= a_{n} =a+( n-1) d

Substitute the value in the above equation

80=5+(n-1)3

80=5+3n-3

80=2+3n

78=3n

n=78/3

n=26

So ,the number of term in given A.P is 26

**Selection of terms concept in arithmetic progressions class 10 **

In some problems, we need to select some terms which are in A.P.So let’s see how we can select terms. Questions are asking by the board on this concept for arithmetic progression class 10

Number of terms terms

3 a-d , a ,a+d

4 a-2d ,a-d ,a+d ,a+2d

5 a-2d ,a-d , a, a+d ,a+2d

In this way, we can select terms in A.P

Don’t worry. We will later understand this concept with examples clearly But here are some property of A.P which you must know

**Property .1**

If a,b and c are three numbers in A.P .then,

2b= a+c

**b=a+c/2**

b is known as the **Arithmetic mean of a and c.**

Let’s take examples to undertsand these concepts clearly

Q.1 If 2x , x+10 , 3x+2 are in A.P .Find the value of x

Sol. Now ,we know if a , b and c are in A.P then ,

2b= a+c

2(x+10)=2x+3x+2

2x+20=5x+2

18=3x

x=6

Q.2 The sum of three numbers in A.P is -3 ,and their product is 8 .Find the numbers

sol. Here is need of concept ” selection of term in A.P “.Now we have to select three numbers which are in A.P

we can select ** a -d , a ,a+d** as three term in A.P

According to question,

The sum of these three number =-3

a -d +a +a+d=-3

3a=-3

a=-1 ………………(i)

since , product of these numbers =8

( a -d ) ( a ) (a+d)=8

( a -d )(a+d)( a )=8

(a²-d²)(a)=8 …….(ii)

substitute the value of a in equation (ii)

{ (-1)²-d²}(-1)=8

{1-d²}(-1) =8

-1+d²=8

d²=9

d=±3

Case-I When a=-1 and d=+3

Numbers will be a -d , a ,a+d = -1-3,-1,-1+3 = -4 ,-1 ,2

Case-II When a=-1 and d=-3

Then , number will be a-d , a ,a+d = -1-(-3) , -1 ,-1-3=2 ,-1 ,-4

Q.3 Determine k so that k²+4k +8 ,2k²+3k+6,3k²+4k+4 are three consecutive terms of the A.P.

Sol. We know that if a,b,c are three consecutive terms of an A.P.then,

2b=a+c

2(2k²+3k+6)=k²+4k +8 +3k²+4k+4

4k²+6k+12=4k²+8k+12

4k²-4k² =8k-6k+12-12

0=2k+0

k=0

### Arithmetic progression class 10 notes [sum of n term ]

This formula is very-very important for class 10 arithmetic progressions .So,always keep in mind

If first term of A.P is a and common difference is d .then,sum of its first n-terms is given by

s_{n}=\frac{n}{2}\{2a+( n-1) d\}

Note= In the sum s_{n} ,Sum of n terms of a sequence is given ,then nth term a_{n} of that sequence can be determined by using the following formula

a_{n}=s_{n} -s_{n-1}

Let’s understand these concepts clearly with the help of examples

Q.1 Find the sum of the 20 terms of the A.P 1,4,7,10……

Sol. In the given A.P 1,4,7,10……

First term(a)=1

Common difference (d)=3

Number of term (n)=20

Substitute these value in the given formula

=n/2{2a+(n-1)d}

=20/2{2(1)+(20-1)3}

=10{2+57}

=10{59}

=590

Q.2 If s_{n} the sum of first n terms of the an A.P is given by S_{n}=5n²+3n,then find its nth term’

sol.Here ,we will use concept of a_{n}=s_{n} -s_{n-1}.let’s see how we can apply

since, S_{n}=5n²+3n ……..(i)

we have to find the value of s_{n-1}.so subtitute (n-1) in place of n in above equation

s_{n-1}=5(n-1)²+3(n-1)

s_{n-1}=5(n²+1-2n)+3n-3

s_{n-1}=5n²+5-10n+3n-3

s_{n-1}=5n²-7n+2 ……….(ii)

Now put value of equation(i) and (ii) in the a_{n}=s_{n} -s_{n-1}.

It will become

a_{n}=(5n²+3n )-(5n²-7n+2)

a_{n}=5n²+3n-5n²+7n-2

a_{n}=10n-2 is the required answer

So,we have discussed about all topic of arithmetic progression class 10 under arithmetic progression class 10 notes.Let’s discuss about ncert solution of A.P

### Arithmetic progression class 10 ncert solution

Here , you shall get ncert solution arithmetic progressions class 10 exercise-wise.

##### Arithmetic progression class 10 ncert solution[exercise 1.1]

Q.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 km when the fare is Rs.15 for the first km and Rs.8 for each additional km

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at the time.

(iii) 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

Sol. (i)

The taxi fare after each km when the fare is Rs15

The fare for each additional km is Rs8

Suppose if a man cover 1 km, the fare will be =8km

If he covers two km =15+8=23

If he covers 3km =15+8+8=31

As we see the fare is increased by the same amount for all km i.e Rs.8

So the given statement will form AP because the common difference is the same .

(ii)Let the initial volume of air in cylinder=V

So,Vaccum pump removes 1/4th of air at the time

The air removed by vacuum pump=1/4×v=v/4

The air remaining in cylinder =V-V/4=3V/4

Again, the Vaccum pump removes 1/4th air=1/4×3V/4=3V/16

Remaining air=3V/4-3V/16=

(iii) Cost of digging for first meter =Rs.150

and Cost increases after every metre=Rs.50

So, Here first term=150 and common difference=50

∴ Given statement forms AP

Q.2 Write first four term of the A.P ,when the first term a and the commin difference d are given as follows:

(i) a=10 ,d=10

(ii) a=-2 ,d=0

(iii) a=4 ,d=-3

(iv) a=-1 ,d=1/2

(v) a=-1.25 ,d=-0.25

sol.(i) we know when a is the first term and d is the common difference of the A.P

A.P will be a, a+d,a+2d,a+3d,a+4d…….

Here a=10 and d=10 ,so A.P will be 10 ,10+10 ,10+2(10),10+3(10)…..

Our first four terms are 10 ,20,30,40.

(ii) Here a=-2 and d=0

so Our formed A.P -2 ,-2+0,-2+2(0)…….

AP will be -2,-2,-2….

(iii) We have a=4 ,d=-3

so ,first four term of A.P is 4,4+(-3),4+2(-3),4+3(-3)=4 ,1,-2 ,-5

(iv) Here a=-1 and d=1/2

so ,first four term will be -1,-1+1/2,-1+2(1/2),-1+3(1/2)=-1,-1/2,0 and 1/2

(v) Here a=-1.25 and d=-0.25

so ,its first four terms are -1.25,-1.25+(-0.25),-1.25+2(-0.25),-1.25+3(-0.25)=-1.25,-1.50,-1.75 and 2

Q.3 For the following AP’s ,write the first term aand the common difference

(i) 3 ,1,-1,-3

(ii) -5 ,-1,3,7

(iii) 1/3,5/3,9/3,13/3….

(iv) 0.6,1.7,2.8,3.9…..

Sol(i) As we know that common diffence is difference bewteen two terms of AP

Here ,second term-first term=1-3=-2

fourth term-third term=-3-(-1)=-2

So from these results ,our first term is 3 and common diffrence is -2

Sol(ii) Here ,second term-first term=-1-(-5)=4

third term-second term=3-(-1)=4

therefore first term is -5 and common diffrence is 4

sol(iii)Here ,second term-first term=5/3-1/3=4/3

third term-second term=9/3-5/3=4/3

Therefore first term =1/3 and common difference =4/3

sol(iv)Here,second term-first term=1.7-0.6=1.1

third term-second term=2.8-1.7=1.1

Therefore ,first term=0.6 and common difference=1.1

Q.4 Which of the following are AP’s ? If they form an AP ,find the common difference d and write three more terms

(i) 2,4,8,16….

(ii) 2,5/2,3,7/2…..

(iii) -1.2 ,-3.2,-5.2,-7.2…….

(iv) -10,-6,-2,2…..

(v) 3,3+√2,3+2√2,3+3√2….

(vi) 0.2 ,0.22,0.222,0.2222….

(vii)0,-4,-8,-12

(viii)-1/2,-1/2,-1/2,-1/2….

(ix)1,3,9,27….

(x)a,2a,3a,4a….

(xi) a,a²,a³…..

(xii)√2,√8,√18,√32…

(xiii) √3,√6,√9,√12…

(xiv) 1²,3²,5²,7²……

(xv)1²,5²,7²,73….

So as we have earlier read that any sequence will be in AP ,If the diffence between all terms are same.

Sol(i)So let’s check diffrence between terms

second term-first term=4-2=2

third term-second term=8-4=4

Here common difference is not same .So this sequence is not in AP

sol(ii) Diffrence between terms

second term-first term=5/2-2=1/2

third term-second term=3-5/2=1/2

fourth term-third term=7/2-3=1/2

since ,common difference of AP is same and equal to 1/2.so this sequence is in AP

sol(iii) Diffrence between terms

second term-first term=-3.2-(-1.2)=-3.2+1.2=-2

third term-second term=-5.2-(-3.2)=-5.2+3.2=-2

fourth term-third term=-7.2-(-5.2)=-2

since ,common difference of AP is same and equal to -2.so this sequence is in AP

sol(iv) Diffrence between terms

second term-first term=-6-(-10)=4

third term-second term=-2-(-6)=4

since ,common difference of AP is same and equal to 4..so this sequence is in AP

sol(v) Diffrence between terms

second term-first term=3+√2-3=√2

third term-second term=3+2√2-(3+√2)=√2

since ,common difference of AP is same and equal to √2.so this sequence is in AP

sol(vi)Here,Diffrence between terms

second term-first term=0.22-0.2=0.02

third term-second term=0.222-0.22= 0.002

Here common difference is not same .So this sequence is not in AP

sol(vii)0,-4,-8,-12…..

Difference between terms

second term-first term=-4-(0)=-4

third term-second term= -8-(-4)=-8+4=-4

fourth term-third term=-12-(-8)=-12+8=-4

since ,common difference of AP is same and equal to -4.so this sequence is in AP

sol(viii)-1/2,-1/2,-1/2,-1/2…..

Difference between terms

second term-first term=-1/2-(-1/2)=0

third term-second term=-1/2-(-1/2)=0

since ,common difference of AP is same and equal to o.So this sequence is in AP

sol(ix) 1,3,9,27……

Difference between terms

second term-first term=3-1=2

third term-second term=9-3=6

Here common difference is not same .So this sequence is not in AP

sol(x) a,2a,3a,4a…….

Difference between terms

second term-first term=2a-a=a

third term-second term=3a-2a=a

since ,common difference of AP is same and equal to a.So this sequence is in AP

sol(xi) a,a²,a³………

Difference between terms

second term-first term=a²- a=a(a-1)

third term-second term=a³-a²=a²(a-1)

Here the common difference is not the same .So this sequence is not in AP

sol(xii) √2,√8,√18,√32…

second term-first term=√8- √2=2√2-√2=√2

Third term-second term=√18-√8=3√2-2√2=√2

since, common difference of AP is the same and equal to a. So this sequence is in AP

sol(xiii) √3,√6,√9,√12…

Second term-first term=√6-√3=√3×√2-√3=√3(√2-1)

Third term-second term=√9-√6=3-√3×√2=√3(√3-√2)

Here common diffrence is not same.So this sequence is not in AP.

sol(xiv) 1²,3²,5²,7²……

Second term-first term=3²-1=9-1=8

Third term-second term=5²-3²=25-9=16

Here common diffrence is not same.So this sequence is not in AP

sol(xv) 1²,5²,7²,73…..

Second term-first term=5²-1=25-1=24

Third term-second term=7²-5²=49-25=24

Fourth term-Third term=73-7²=73-49=24

since ,common difference of AP is same and equal to 24.So this sequence is in AP

##### Arithmetic progression class 10 ncert solution[exercise 5.2]

Here ,you will get complete question solution of arithmetic progression class 10 exercise 5.2

Q.1 Fill in the blanks in the following table,given that a is the first term,d is a common difference and a_{n} the nth term of the A.P

Q.2 Choose the correct choice in the following and justify :

(i) 30th term of the AP :10,7,4,……..is

(A) 97 (B) 77 (C)-77 (D) -87

Sol(i) We have AP as : 10,7,4,…….

Here first term =10,common difference=-3 and n=30

we know a_{n} =a+( n-1) d

a_{30} =10+(30-1)(-3) a_{30}= 10+(29)(-3) a_{30}=10-87 a_{30}=-77The correct option is (C)=-77

(ii) 11th term of the A.P : -3 ,-1/2 ,2,…..is

(A) 28 (B) 22 (C) -38 (D)-481/2

sol(ii) We have AP as -3 ,-1/2 ,2,….

Here first term(a)=-3 ,common diffrence=-1/2-(-3)=-1/2+3=5/2 and n=11

we know a_{n} =a+( n-1) d

a_{11}=-3+(11-1)5/2 a_{11}=-3+10(5/2) a_{11}= -3+25 a_{11}=22The correct option is (B)=22

Q.4 Which term of the AP : 3,8,13,18……is 78?

Sol.Let the nth term of AP will be 78

Here first term(a)=3 ,common difference(d)=5 and a_{n}=78

we know , a_{n} =a+( n-1) d

78=3+(n-1)5

78=3+5n-5

78=-2+5n

5n=80

n=16

so 16 th term of AP is 78

Q.5 Find the number terms in the each of the following A.P’s :

(i) 7,13,19,…….205 (ii) 18,151/2,13………-47

sol(i) We have 7,13,19,…….205 as a AP

Here ,first term(a)=7 ,common difference(d)=6 and last term(l)=205

we have read about formula for last term

l=a+( n-1) d

205=7+(n-1)6

205=7+6n-6

205=1+6n

6n=204

n=34

So, this AP has 34 terms

sol(ii) we have 18,151/2,13………-47

Here .first term(a)=18 ,common difference(d)=151/2-18=-5/2 and last term(l)=-47

we know

l=a+( n-1) d

-47=18+(n-1)(-5/2)

-65=-5n/2+5/2

5n/2=5/2+65

2.5n=67.5

n=27

Q.6 Check wheather -150 is a term of the AP : 11 ,8,5,2…

Sol.Let -150 is nth term of given AP

Here first term(a)=11 and common difference(d)=8-11=-3

We know a_{n} =a+( n-1) d

-150=11+(n-1)(-3)

-150=11-3n+3

-150=14-3n

-150-14=-3n

-164=-3n

n=164/3

Here value of n is fractional .so it cannot be term of given AP

Q.7 Find the 31 st term of the AP whose 11 th term is 38 and the 16th term is 73.

sol. Let first term and common difference of AP is a and d respectively

since it is given a_{11}=38 and a_{16}=73

we can write a_{11}=a+10d and a_{16}=a+15d

It becomes

a+10d=38 …..(i)

a+15d=73 ……(ii)

subtract equation(i) and (ii)

(a+10d)-( a+15d)=38-73

a +10d-a-15d=-35

-5d=-35

d=7

subtitute this value in equation (i)

a+10d=38

a+10(7)=38

a=38-70

a=-32

Therefore 34th term will be a+33d

a_[34}=a+30d

a_{34}=-32+7(30)

a_{34}=-32+210

a_{34}=178

Hence 34th term of AP is 178

Q.8 An AP consists of 50 terms of which 3 rd term is 12 and the last term is 106 .Find the 29 th term term.

sol. Let first term and common difference of AP is a and d respectively

Number of term in AP (n)=50

third term=12

last term(a_{n})=106

we know a_{n} =a+( n-1) d

106=a+(50-1)d

106=a+50d-d

106=a+49d ……(i)

since it is also given third term=12

a+2d=12 ….(ii)

subtract equation (i) and (ii)

(a+2d)-(a+49d)=12-106

-47d=-94

d=2

subtitute this value in equation (ii)

a+2d=12

a+2(2)=12

a+4=12

a=8

Now we have first term and common difference as 8 and 2 .we can easily find 29th term

since a_{29]=a+28d

a_{29]=8+28(2)

a_{29]=8+56

a_{29]=64

Q.9 If the 3rd and the 9th term of an A.P are 4 and -8 respctively ,which term of the this AP is Zero .

Sol. Let first term and common difference of AP is a and d respectively

It is given a_{3}=4 ,a_{9}=-8

a_{3}=a+2d , a_{9}=a+8d

it becomes

a+2d =4 ……(i)

a+8d=-8

solving these equations we get,a=8 and d=-2

Now we have to which term of this AP is zero

So , let n th term of AP is o

we know a_{n} =a+( n-1) d

0=8+(n-1)(-2)

0=8-2n+2

2n=10

n=5

Therefore 5 th term of given AP is 0

Q.10 The 17 th term of an AP exceeds its 10 term by .Find the common difference.

sol.Let first term and common difference of AP is a and d respectively

so we can write a_{17}=a+16d and a_{10)=a+9d

According to question,17 th term of an AP exceeds its 10 term by7

a+16d=a+9d+7

7d=7

d=1

Hence common diffrence of AP is 1

Q.11 Which term of the AP :3,15,27,39….will be 132 more than its 54th term?

sol.We have 3,15,27,39… as a AP

here first term(a)=3 and common difference(d)=12

Now its 54th term will be ‘

a_{54}=a+53d

a_{54}=3+53(12)

a_{54}=3+636

a_{54}=639

Now we have to find about the term which is 132 more than 54th term

So,required number will be 639+132=771

Let it is nth term of AP

a+(n-1)d=771

3+(n-1)12=771

3+12n-12=771

12n=771+9

12n=780

n=65

so 771 is 65 th term of given AP

Q.13 How many three -digits numbers are divisible by 7?

sol.Firstthree -digits number which is exactly divisible by7 is 105

secon number ,third ,fourth are 112,119,126

As we know largest 3-digit number is 999

but when we divide 999 by 7 ,we get 5 as remainder

That’s means 999-5=994 is the largest 3 -digit number which is exactly by 7

Required AP is 112,119,126……..994

Here we have First term (a)=112 ,common diffrence(d)=7 and last term(a_{n})=994

we know

a_{n} =a+( n-1) d994=112+(n-1)7

994=112+7n-7

994=105+7n

994-105=7n

889=7n

n=127

Hence ,there are 127 three-digit number which is exactly divisible by 7

Q.14 How many multiples of 4 lie between 10 and 250?

sol.

As we know ,12 is first number after 10 which is exactly divisible by 4

and when we 250 by 4 we get 2 as remainders i.e 248 is the last term between 250 which is exactly divisible by 4

Since we have first term as 12 ,common difference =4 and last term is 248

so Our AP will be : 12,16,20……….248

we know last term=a+(n-1)d

248=12+(n-1)4

248=12+4n-4

248=8+4n

240=4n

n=60

Hence there are 60 multiples between 10 and 250.

Q.15 For what value of n , re the nth terms of the two AP’s:63,65,67….and 3,10,17…..equal?

sol.

We have the t term and the on difference of both AP’s are 63 ,3 and 3,7 respectively.

a=63 ,d= and a_{1}=3 ,d_{1}=7

A/Q

a+(n-1)d=a_{1}+(n-1)d_{1}

63+(n-1)2=3+(n-1)7

63+2n-2=3+7n-7

61+2n=7n-4

65=5n

n=13

so 13th term of both AP are same

Q.16 Determine the A.P whose third term is 16 and the 7th term exceeds the 5 th term by the 12.

sol.Let the first term and common difference of AP are a and d respectively

It is given a_{3}=16 and a_{7}=a_{5}+12

a+2d=16 , a+6d=a+4d+12

a+2d=16 , 2d=12

a+2d=16 , d=6

subtitute d=6 in equation in a+2d=16

a+2d=16

a+2(6)=16

a=16-12

a=4

since we have first term and common difference as 4 and 6.so,AP will be 4,10,16,22….

Q.17 Find the 20th etrm from the last term of the AP :3,8,13…….253.

sol.Now here is little twist ,twist is we have to 20th term but end not beginning

so simply reverse the given AP

253, 248, 243,…………….13,8,5

From this AP we have first term=253 and common diffrence=-6

we know a_{n} =a+( n-1) d

a_{20}=253+(20-1)(-5)

a_{20}=253+19(-5)

a_{20}=253-95

a_{20}=158

Q.18 The sum of the 4th term and 8th term of an AP is 24 and the sum of the 6th and 10th term is 44.Find the first three terms of th AP

sol.Let the first term and common difference of AP are a and d respectively

It is given that

a_{4}+a_{8}=24 ,a_{6}+a_{10}=44

a+3d+a+7d=24 ,a+5d+a+9d=44

2a+10d=24 …(i) ,2a+14d=44……(ii)

subtract equation (i) and (ii)

-4d=-20

d=5

subtitute d=5 in equation (i)

2a+10d=24

2a+10(5)=24

2a=24-50

2a=-26

a=-13

Now we have first term=-13 and common difference=5

Required AP =-13,-8–3….

Q.19 Subha Rao started work in 1995 at the annual salary of Rs 5000 and received an increment of Rs 200 each year .In which year did his income reach Rs 7000?

sol.Subha Rao starting salary =5000 i.e first term=5000

The increment in her salary per year =200 i.e common diffrence=200

Subha salary is now=7000 i.e last term=7000

so, a_{n} =a+( n-1) d

7000=5000+(n-1)200

2000=200n-200

2200=200n

n=11

Subha salary will 7000 after year

the year will be 1995+11=2006

Q.20 Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs.1.75.If in the nth week, her weekly savings become Rs.20.75.Find the value of n?

sol. Ramkali saved in the first week=Rs.5

His week saving increase by 1.7

Her saving nowadays=20.75

a_{n} =a+( n-1) d

20.75=5+(n-1)(1.75)

20.75=5+1.75n-1.75

20.75=3.25+1.75n

17.5=1.75n

n=10

##### Arithmetic progression class 10 ncert solution[exercise 5.3]

Here ,you will get arithmetic progressions class 10 exercise 5.3 solution

Q.1 Find the sum of the following APs:

(i) 2,7,12,……….to 10 terms.

(ii) -37,-33,-29……to 12 terms

(iii) 0.6 ,1.7,2.8……to 100 terms

(iv)1/15,1/12,1/10……to 11 terms

sol(i)2,7,12,……….to 10 terms.

Here a=2 ,d=5 and n=10

we know

s_{n}=\frac{n}{2}\{2a+( n-1) d\}

s_{10}=10/2{2(2)+(10-1)5

s_{10}=5{4+45}

[s_{10}=5{49}

s_{10}=245

sol(ii)-37,-33,-29……to 12 terms

a=-37 ,d=-33-(-37)=4 and n=12

we know

s_{n}=\frac{n}{2}\{2a+( n-1) d\}

s_{12}=12/2{2(-37)+(12-1)4}

s_{12}}=6{-74+44}

s_{12}=6{-30}

s_{12}=-180

sol(iii)0.6 ,1.7,2.8……to 100 terms

a=0.6 ,d=1.7-0.6=1.1 and n=100

we know

s_{n}=\frac{n}{2}\{2a+( n-1) d\}

s_{100}=100/2{2(0.6)+(100-1)(1.1)

s_{100}=50{1.2+108.9}

s_{100}=50{110.1}

s_{100}=5505

sol(iv) 1/15,1/12,1/10……to 11 terms

a=1/15 ,d=1/12-1/15=1/60 and n=11

we know ,

s_{n}=\frac{n}{2}\{2a+( n-1) d\}

s_{11}=11/2{2(1/15)+(11-1)1/60

s_{11}=11/2{2/15+10/60}

s_{11}=11/2{9/30}

s_{11}=33/20

Q.2 Find the sums given below:

(i) 7+10.1/2+14+…….84.

sol.we have a=7 ,d=10.1/2-7=7/2 and last term=84

before finding its sum ,we must know about number of term in AP

since, l=+n-1)d

84=7+(n-1)7/2

77=(n-1)7/2

11=(n-1)/2

22=(n-1)

n=23

we use formula whenever last term of AP is given

s_{n}=n/2{a+l}

s_{23}=23/2{7+84}

s_{23}=23/2{91}

s_{23}=2093/2

(ii) 34+32+30+……..10

we will apply same procedure here as well

a=34 ,d=-2 and last term=10

so ,l=a+(n-1)d

10=34+(n-1)(-2)

-24=(n-10)(-2)

12=(n-10)

n=22

s_{n}=n/2{a+l}

s_{22}=22/2{34+10}

s_{22}=11{44}

s_{22}=484

(iii)-5+(-8)+(-11)………..+(-230)

sol(iii) we have a=-5 ,d=-8-(-5)=-8+5=-3 and last term(l)=-230

l=a+(n-1)d

-230=-5+(n-1)(-3)

-230=-5-3n+3

-230=-2-3n

-228=-3n

n=76

s_{n}=n/2{a+l}

s_{n}=76/2{-5-230}

s_{n}=38{-235}

s_{n}=8930

Q.3 In the AP

(i) given a=5 ,d=3,a_{n}=50,find n and s_{n}.

sol. Given that :

a=5 ,d=3 and a_{n}=50

we know

a_{n}=a+(n-1)d

50=5+(n-1)3

50=5+3n-3

50=2+3n

48=3n

n=16

Now we have a,d and n.we can easily find its sum

s_{n}=n/2{a+l}

s_{n}=16/2{5+50}

s_{n}=8{55}

s_{n}=440

(ii) given a=7 ,a_{13}=35,find d and s_{13}

sol.

Given that :

a=7 ,a_{13}=35

we write a_{13}=a+12d

35=7+12d

28=12d

d=2

So,the common diffrence is 2

Sum of 13 term will be

s_{n}=n/2{a+l}

s_{13}=13/2{7+35}

s_{13}=13/2{42}

s_{13}=13×21

s_{13}=273

(iii)given a_{12}=37 ,d=3,find a and s_{12}

sol.

It is given that

a_{12}=37 and d=3

a_{12}=a+11d

37=a+11(3)

37=a+33

a=5

(iv)given a_{3}=15,s_{10}=125,find d and a_{10}

sol. Let first term and common diffrence of AP are a and d respectively

It is given that

a_{3}[.katex]=15</p> <p>a+2d=15 ........(i)</p> <p>[katex]s_{10}=125

10/2{2a+(10-1)d}=125

5{2a+9d}=125

2a+9d=25 ..........(ii)

Multiply equation (i) by 2 and subtract

2(a+2d)=2(15)

2a+4d=30 ......(iii)

(2a+4d)-(2a+9d)=30-25

-5d=5

d=-1

subtitute d=-1 in equation (i)

a+2d=15

a+2(-1)=15

a-2=15

a=17

so,the first term and common diffrence of AP are 17 and -1 respectively

Its a_{10} will be

a_{10}=a+8d

a_{10}=17+8(-1)

a_{10}=17-8

a_{10}=9

(v) given d=5 ,s_{9}=75,find d and a_{10}

sol.

Given:

d=5 and s_{9}=75

we know

s_{n}=\frac{n}{2}\{2a+( n-1) d\}

75=9/2{2a+(9-1)5}

75=9/2{2a+40}

75=9/2.2{a+20}

75=9{a+20}

75=9a+180

9a=180-75

9a=105

a=105/9

a=35/3

Now we can easily find the valuea_{10}

a_{10}=a+9d

a_{10}=35/3+9(5)

a_{10}=35/3+45

a_{10}=56.6

Q.4 How many terms of the AP:9,17,25........must be taken to give a sum of 636?

sol.We have AP as : 9,17,25......

here a=9 ,d=17-9=8 and s_{n}=636

we have to find n

s_{n}=\frac{n}{2}\{2a+( n-1) d\}

636=n/2{2(9)+(n-1)8}

636=n/2{18+8n-8}

636=n/2{10+8n}

1272=n{10+8n}

1272=10n+8n²

2(636)=2(5n+4n²)

5n+4n²-636=0

4n²+53n-48n-636=0

n(4n+53)-12(4n+53)=0

(4n+53)(n-12)=0

4n+53=0 ,n-12=0

n=-53/4 n=12

fractional value is not possible .So answer is 12

number of term =12

Q.5 The first term fof an AP is 5,the last term is 45 and the sum is 400.Find the number of terms and the common difference.

sol.Let common difference and number of term in AP are d and n respectively.

Given:

first term(a)=5

last term(l)=45

sum of 45 terms (s_{45})=400

we know

s_{n}=n/2{a+l}

400=n/2{5+45}

400=n/2{50}

400=25n

n=16

Now we have to find common diffrence (d)

l=a+(n-1)d

45=5+(16-1)d

40=15d

d=8/3

common difference of AP=8/3

Q.6 The first and the last terms of an AP are 17 and 350 respectively.If the common diffrence is 9 , ow many terms are there,and what is their sum?

sol. we have

first term(a)=17

common diffrence(d)=9

last term(l)=350

we know

l=a+(n-1)d

350=17+(n-1)9

333=(n-1)9

37=(n-1)

n=38

Sum of 38 terms will be

s_{n}=\frac{n}{2}\{2a+( n-1) d\}

s_{38}=38/2{2(17)+(38-1)9

s_{38}=19{34+333}

s_{38}=19{367}

s_{38}=6973

Q.7 Find the sum of first 22 terms of an Arithmetic progression in which d=7 and 22nd term is 149.

sol.Let first term fo AP is a

given:

a_{22}=149

a+21d=149

a+21(7)=149

a+147=149

a=149-147

a=2

sum of 22 term will be

s_{n}=\frac{n}{2}\{2a+( n-1) d\}

s_{22}=22/2{2(2)+(22-1)7

s_{22}=11{4+147}

s_{22}=11{151}

s_{22}=1661

Q.8 Find the sum of first 51 terms of an AP whose second and the third terms are 14 and 18 respectively.

sol.Let first term and common diffrence of AP are a and d respectively

Given

a_{2}=14 , a_{3}=18

a+d=14 ....(i) a+2d=18 .....(ii)

subtract equation (i) and (ii)

-d=-4

d=4

subtitute d=4 in equation (i)

a+d=14

a+4=14

a=10

sum of 51 term will be

s_{n}=\frac{n}{2}\{2a+( n-1) d\}

s_{51}=51/2{2(10)+(51-1)4

s_{51}=51/2{20+200}

s_{51}=51/2{220}

s_{51}=51×110

s_{51}=5610

Q.9 If the sum of the first 7 terms of an AP and that of 17 terms is 289,find the sum of first n terms

sol.Let first term and common diffrence of AP are a and d respectively

Given:

s_{7}=49</p> <p>s_{17}=289</p> <p>it becomes</p> <p>7/2{2a+(7-1)d}=49 ........(i)</p> <p>17/2{2a+(17-1)d}=289 .........(ii)</p> <p>2a+</p> <p>Q.10 Show that a_{1},a_{2} ,......a_{n}.....form an AP where a_{n} is defined as below:</p> <p>(i) a_{n}=3+4n (ii) a_{n}=9-5n</p> <p>sol(i)a_{n}=3+4n</p> <p>Find value a_{1},a_{2},a_{3},a_{4}</p> <p>a_{1}=3+4(1)=7</p> <p>a_{2}=3+4(2)=11</p> <p>a_{3}=3+4(3)=15</p> <p>difference=a_{2}-a_{1}=11-7=4</p> <p>difference=a_{3}-a_{2}=15-11=4</p> <p>Its difference is same and equal to 4.so this sequence form an AP</p> <p>The sum of first term of this AP will be</p> [katex] s_{n=}\frac{n}{2}\{2a+( n-1) d\}

s_{15}=15/2{2(7)+(15-1)4}

s_{15}=15/2{14+56}

s_{15}=15/2{70}

s_{15}=525

sol(ii) a_{n}=9-5n

a_{1}=9-5(1)

a_{1}=4

a_{2}=9-5(2)

a_{2}=-1

a_{3}=9-5(3)

a_{3}=9-15

a_{3}=-6

So

difference=a_{2}-a_{1}=(-1)-4=-5

difference=a_{3}-a_{2}=(-6)-(-1)=-6+1=-5

Since common diffrence of this sequence is same and equat to -5.so this is a AP

Now sum of first 15 term will be

s_{n=}\frac{n}{2}\{2a+( n-1) d\}s_{15}=15/2{2(4)+(15-1)(-5)}

s_{15}=15/2{8-70}

s_{15}=15/2{-62}

s_{15}=15×(-31)

s_{15}=-465

Also,find the sum of the first 15 terms in each case.

Q.11 If the sum of the first n terms of an AP is 4n-n²,what is the first term(that is s_{1})?What is the sum of first two terms ? What is the second term ?similarly ,find the 3rd,the 10th and the nth terms.

sol.

Q.12 Find the sum of the first 40 positive integers divisible by 6.

sol.The positive integer that is divisible by 6 i.e 6,12,18,24...

From above sequence ,we have first term=6 and common diffrence=6

The sum of first 40 positive integer will be

s_{n=}\frac{n}{2}\{2a+( n-1) d\}s_{40}=40/2{2(6)+(40-1)6}

=20{12+234}

=20{246}

=4920

Q.13 Find the sum of the first 15 multiples of 8.

Sol.The multiples of 8 are 8,16,24,32.......

From this sequence we have ,

a=8 and d=8

So the sum of first 15 term will be given by

s_{n=}\frac{n}{2}\{2a+( n-1) d\}s_{15}=15/2{2(8)+(15-1)8}

s_{15}=15/2{16+112}

s_{15}=15/2{128}

s_{15}=15×64

s_{15}=960

Q.14 Find the sum of the odd numbers between 0 and 50.

sol.Odd numbers between 0 and 50 are 3,5,7,.......49

From this sequence we have ,

a=3 ,d=5-3=2 and last term=49

we know

last term=a+(n-1)d

49=3+(n-1)(2)

49=3+2n-2

49=1+2n

48=2n

n=24

There are 24 odd terms between 0and 50

Sum of 24 term will be

s_{n=}\frac{n}{2}\{2a+( n-1) d\}s_{24}=24/2{2(3)+(24-1)2}

s_{24}=12{6+46}

s_{24}=12{52}

s_{24}=624

Sum of 24 term will be 624

Q.15 A contract on construction job specifies a penalty for delay of the completion beyond a certain date as follows :Rs 200 for the first day ,Rs250 for the second day ,Rs 300 for the third day,etc .the penalty for each succeding day being 50 more than for the preceding day.How much money the contracter has to pay as penalty ,if he has delayed the work by30 days?

Sol. If we carefully read the problem ,we can guess penalties are in AP with first term=200 and common difference=50

Amount of money contractor have to when he delay the work by 30 days is given

s_{n}=\frac{n}{2}\{2a+( n-1) d\}s_{n}=20/2{2(200)+(20-1)(50)}

s_{n}=10{400+950}

s_{n}=10{1350}

s_{n}=13500

Q.16 A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance .If the each prize is Rs 20 less than its preceding prize ,find the value of each of the prizes.

Sol.According to question ,the value of prize decrease by 20 Rs with each prize

Let the value of first price =Rs.p

then,the value of secon price =(p-20)

third price =(p-20-20)=(p-40)

As we see ,this sequence seems to be AP will first term=p and common diffrence=-20 (not 20 )

now ,we have sum of total seven prize=700

s_{7}=700

7/2{2p+(7-1)(-20)}=700

7/2{2p-120}=700

7{2p-120}=1400

(2p-120)=200

2p=320

p=160

Value of all prizes are Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40

Q.17 In a school ,students thought of planting trees in and around the school to reduce air pollution .It was decided that the number of trees ,that each section of each class will plant .will be the same as the class ,in which they are studying .e.g a section of class 1 will plant 1 tree ,a section of class II will plant 2 trees and so on till class XII .There are three sections of each class.How many tress will be planted by the students?

Sol. If we carefully read the problem , we get number of trees planted by students are in AP

1,2,3,4,...........12

First term=1 ,common difference=1 and n=12

s_{12}=n/2{2a+(n-1)d}

s_{12}=12/2{2(1)+(12-1)(1)}

s_{12}=6{2+11}

s_{12}=78

Q.18 A spiral is made up of successive semicircles ,with centres alternately at A and B,starting with cente A,of radii 0.5cm ,1.0cm,1.5cm,2.0cm,......as shown figure in book .What is the total length of such spiral made up of thirteen consecutive semicircles?(Π=22/7)

Sol.

Q.19 200 logs are stacked in the following manner: 20 logs in the bottom row ,19 in the next row ,18 in the row next to it and so on .....In how many rows are the 200 logs placed and how many logs are in the top row?

Sol.Total number of logs =200

Logs are in such a manner , they are forming Arithmetic progression

20,19,18,17.......

Now we have to find number of rows in which all thse logs placed completely and number of log in top row (last row)

Let there are n number of rows

S_{n}=200

n/2{2a+(n-1)d}=200

n/2{2(20)+(n-1)(-1)}=200

400=n(41-n)

400=41n-n²

n²-41n+400=0

n²-16n-25n+400=0

n(n-16)-25(n-16)=0

(n-16)(n-25)=0

n=16 or n=25

Number of logs in top row are given by

a_{n}=a+(n-1)d

a_{}=20+(16-1)(-1)

a_{}=20-15

a_{16}=5

but when we use n=25

A_(25)=20+(25-1)(-1)

a_{25}=20-24

a_{25}=-4

So ,here number of logs in last row are negative

therefore n=25 cannot be possible

Q.20 In the potatoes race , a bucket is placed at the starting point ,which is 5 m from the first potato,and the other potatoes are placed 3m apart in straight line.There are ten potatoes in the line(see figure in book)

A competitor starts from the bucket ,pick up the nearest potato ,runs back with it ,drops it in bucket ,runs back to pick up thr next potato ,runs to the bucket to drop it in ,and she continies in the same way until all potatoes are in the bucket.What is the total distance the competitor has to run?

Sol.

Distance run by the competitor to pick up the first potato=2×5m

Distance run by the competitor to pick up the second potato=2(5+3)m

Distance run by the competitor to pick up the third potato =2(5+2×3)m

.

.

Distance run by the competitor to pick up the nth potatoes=2{5+(n-1)×3}m

Total distance covered by the competitor =d_{1}+d_{2}+d_{3}+........d_{n}

=

##### Arithmetic progression class 10 ncert solution[exercise 5.4](optional)

Here are optional exercise question solution of class 10 arithmetic progression

Q.1 Which term of the AP:121,117,113.......is its first negative term?

Sol.I tell you ,a simple hack to solve this types of problem.

Always remember that this type of AP are always decreasing

so, instead of finding first term find the term which is zero because after term after it will negative

Here we have first term=121 ,common diffrence=117-121=-4

Let its nth term is 0

a+(n-1)d=0

121+(n-1)(-4)=0

121-4n+4=0

125-4n=0

4n=125

n=125/4

n=31.25

That means our 32 th term will be first negative term.

Q.2 The sum of the third and the seventh terms of an AP is 6 and their product is 8.Find the sum of first sixteen terms of the AP.

Sol.Let first term and common diffrence of AP are a and d respectively

It is given that

a_{3}+a_{7}=6

a+2d+a+6d=6

2a+8d=6

a+4d=3 ...........(i)

Product of these terms =8

(a+2d)(a+6d)=8..........(ii)

Subtitute a=3-4d from equation (i) in equation (ii)

(3-4d+2d)(3-4d+6d)=8

(3-2d)(3+2d)=8

9-4d²=8

-4d²=-1

d²=1

d=±1/2

Case-I when d=1/2

a+4d=3

a+4(1/2)=3

a=1

The value of a=1

Case-II when d=-1/2

a+4d=3

a+4(-1/2)=3

a-2=3

a=5

The value of a will be 5 when d=-1/2

Now we have two AP's with first term and common diffrence are 1,1/2 and 5,-1/2 respectively

Sum of first sixteen term will be

Case-I When a=1 and d=1/2

s_{n}=\frac{n}{2}\{2a+( n-1) d\}s_{16}=16/2{2(1)+(16-1)(1/2)}

s_{16}=8{2+15/2}

s_{16}=76

s_{16}=76

Case-II when a=5 and d=-1/2

s_{n}=\frac{n}{2}\{2a+( n-1) d\}s_{16}=16/2{2(5)+(16-1)(-1/2)}

s_{16}=8{10-15/2}

s_{16}=20

Q.3 A ladder has rungs 25 cm apart(see figure in book).The rungs decrease uniformly in length from 45cm at the bottom to 25 cm at the top.If the top and the bottom rungs are 2.1/2 m apart.What is the length of the wood required for the rungs?

Sol.

Distance between each ladder=25cm

The length of the bottom and top rungs are 45 and 25 cm

Distance between top and bottom rungs are

Q.4 The houses of a row are numbered consecutively from 1 to 49.Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it .Find this value of x

Sol.Let there be a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it.

1+2+3+.........(x-1)=(x+1)+(x+2)........49

1+2+3+..........(x-1)={1+2+3+4......+x+(x+1)+...49}-{1+2+3+.......x)

x-1/2{1+(x-1}=49/2{1+49}-x/2{1+x}

x(x-1)/2=49×50/2-x(x+1)/2

(x²-x)=49×50-x(x+1)

(x²-x)+x²+x=49×50

2x²=49×50

x²=49×25

x=7×5

x=35

Q.5 A small terrace at a football ground comprises of 15 steps each of which is 50m long and built of solid concrete.

Each step has a rise of 1/4m and a tread of 1/2m(see figure in book).Calculate the total valume of concrete required to build the terrace.

Sol.Length and width of each step are 50 m and 1/2 m respectively

Height of each step=1/4

Height of second step={1/4+1/4}=2×1/4=1/2

Height of third step=3/4 and so on

As we know each step is in form of cuboid

so volume of each step will given by length×breadth×height

Volume of step 1 =(50×1/2×1/4)m³

Volume of step 2 =(50×1/2×1/2}m³ so on

Amount of concrete used in ladder ={50×1/2×1/4}m³+{50×1/2×2(1/4)}m³..........+{50×1/2×15/4}m³

=50×1/2{1/4+2/4+3/4......15/4}

=25{1/4+2/4+3/4......15/4}

=25/4{1+2+3+4......15}

=25/4×15/2(1+15)m³

=25/4×15/2×16m³

=750m³

### Frequently asked question for arithmetic progression class 10 notes

**Q.1 What is the progression?**

Ans. It is a collection of numbers which are arranged according to some definite rule.

Examples:

1. 2,4,6,8,10.............

2. 2,4,8,16,32.........

3.1/3,1/5,1/7.......

**Q.2 What are types of progression?**

Ans.Progression are generally of three types

1.Arithmetic progression

2.Geometric progression

3.Harmonic progression

**Arithmetic progression: The progression in which the difference between consecutive terms are always constant is called an arithmetic progression.**

example:2,4,6,8,10.............

**Geometric progression**: The progression in which the ratio of a term to the preceding term to it is always constant.

example: 2,4,8,16,32.........

**Harmonic progression:The progression which is form through reciprocal of arithmetic progression .**

example: 1/3,1/5, 1/7,1/9 ..........

**Q.3 Who is the father of progression?**

Ans: Carl Friedrich guess is known as the father of progression

**Q4What is the formula for sum of arithmetic progressions?**

Ans. The formula to calculate sum of arithmetic progressions is

s_{n=}\frac{n}{2}\{2a+( n-1) d\}where a and d are the first term and common difference of the A.P and n is the number of term which sum we have to find.

I hope this article [Arithmetic progression class 10 notes ,ncert solution ] are helpful .COMMENT BELOW if you have any doubt regarding any topic .