Class 10 - Chapter 5 Archives – Mathemafia http://sh016.global.temp.domains/~mathemaf/category/class-10-chapter-5/ Mathematics is like music - Enjoy it!!! Fri, 09 May 2025 10:33:20 +0000 en-US hourly 1 https://wordpress.org/?v=7.0 https://mathemafia.com/wp-content/uploads/2020/08/cropped-M__Logo__SM-1-32x32.png Class 10 - Chapter 5 Archives – Mathemafia http://sh016.global.temp.domains/~mathemaf/category/class-10-chapter-5/ 32 32 Exercise 5.2 (NCERT) – Arithmetic Progressions, Class 10 – All Solutions https://mathemafia.com/exercise-5-2-ncert-arithmetic-progressions-class-10-all-solutions/ Fri, 09 May 2025 10:28:18 +0000 https://mathemafia.com/?p=18908 Exercise 5.2, chapter 5, Arithmetic Progressions, Class 10 - This exercise is primarily about finding the nth term of an AP series. Few questions are about finding first term and common difference as well.

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Exercise 5.2 (NCERT) – Arithmetic Progressions, Class 10

Below are the Quick links for all questions of Exercise 5.2, Arithmetic Progressions, Class 10. Click a link to view the solution of corresponding question.

Chapter 5 Exercise 5.2 Class 10

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Below are the solutions of every question of exercise 5.2Β of chapter 5, Arithmetic Progressions, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.Β Β 

Question 1

Fill in the blanks in the following table, given that a is the first term, d the common difference and a the nth term of the AP:

(i) a=7, d=3, n=8, an=?

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nth term of an AP

(ii) a=-18, d=?, n=10, an=0

(iii) a=?, d=-3, n=18, an=-5

(iv) a=-18.9, d=2.5, n=?, an=3.6

(v) a=3.5, d=0, n=105, an=?

Question 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

First Term, a = 10

Common Difference, d = 7 – 10 = -3

n = 30

nth terms of an AP

Answer: C

(ii) 11th term of the AP: – 3, -1/2, 2, … isΒ 

(A) 28 (B) 22 (C) -38 (D) -48 + 1/2

Answer: B

Question 3

In the following APs, find the missing terms in the boxes :

(i) 2, ? , 26

Given AP: 2,Β  ? , 26Β 
Middle term is the mean of adjacent terms in an AP series.Β 

Answer: 14

(ii) ? , 13, ?, 3

Let us write the viven AP as : x, 13, y, 3

Middle term is the mean of adjacent terms in an AP series.Β 

Now the series is : x, 13, 8, 3

Now 13 is mean of x and 8.

Answer: 18, 8

(iii)Β  5, ?, ?, 19/2

First term, a = 5

Let the common difference = d

nth terms of an AP

Let us now find the missing termsΒ 

Answer: 13/2, 8

(iv)Β  -4, ?, ?, ?, ?, 6

First term, a = -4

Let the common difference = d

nth terms of an AP

Let us now find the missing termsΒ 

Answer: -2, 0, 2, 4

(v)Β  ?, 38, ?, ?, ?, -22

Let the First term = a

Let the common difference = d

nth terms of an AP

Subtract (2) from (1)

Substitute value of d in (2)

Let us now find the missing termsΒ 

Answer: 53, 23, 8, -7

Question 4

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

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First term (a) = 3
Common difference (d) = 8 – 3 = 5

nth terms of an AP

Answer: 16th term

Question 5

Find the number of terms in each of the following APs :

(i) 7, 13, 19, . . . , 205

First term (a) = 7
Common difference (d) = aβ‚‚ – a₁ = 13 – 7 = 6

nth terms of an AP

This means

Answer: n = 34

(ii) 18, 15+1/2, 13, . . . , – 47

First term (a) = 18

Common differenceΒ 

nth terms of an AP

This means

Answer: n = 27

Question 6

Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .

First term (a) = 11
Common difference (d) = aβ‚‚ – a₁ = 8 – 11 = -3

nth terms of an AP

This means

Since n is not an integer this implies -150 is not a term of the AP.

Answer: -150 is not a term of given AP

Question 7

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

nth terms of an AP

This means

Subtract (1) from (2)

Substitute value of d in (1)

Let’s find 31st term.

Answer: a31 = 178

Question 8

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

nth terms of an AP

This means

Subtract (1) from (2)

Substitute value of d in (1)

Let’s find 29th term.

Answer: a29 = 64

Question 9

If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

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nth terms of an AP

This means

Subtract (1) from (2)

Substitute value of d in (1)

Let nth term is 0, this implies

Answer: 5th term

Question 10

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

nth terms of an AP

This means

Answer: d = 1

Question 11

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

nth terms of an AP

Also given

Let kth term is 132 more than 54th term, this implies

Answer: 65th term

Question 12

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Let the two APs be

First AP: a1, a2, a3, …Β 

Second AP: b1, b2, b3, …

Let Common difference for both APs = d

nth terms of the APs will be

Difference between 100th terms = 100

Let’s find difference between 1000th terms

Answer: 100

Question 13

How many three-digit numbers are divisible by 7?

Three-digit numbers range from 100 to 999

The smallest 3-digit number divisible by 7 = 105

The largest 3-digit number divisible by 7 = 994Β 

All 3-digit numbers divisible by 7 are:Β 

105, 112, 119, ….. , 994

This is an AP series.

First term, a = 105

Common Difference, d = 7

Let the number of terms in this AP = n

We know that, nth terms of an AP

This implies

Answer: 128

Question 14

How many multiples of 4 lie between 10 and 250?

Smallest multiple of 4 > 10 = 12

Largest multiple of 4 < 250 = 248

All multiples of 4 between numbers between 10 and 250 are:

12, 16, 20, …., 248

This is an AP series.

First term, a = 12

Common Difference, d = 4

Let the number of terms in this AP = n

We know that, nth terms of an AP

This implies

Answer: 60

Question 15

For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

First AP: 63, 65, 67, …

Second AP: 3, 10, 17, …

nth Terms of both APs are equal, this implies

Answer: 13

Question 16

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Let the AP be

a, a + d, a + 2d, a + 3d,…

3rd term = 16

7th term – 5th term = 12

Substitute the value of d in (1)

AP series:

a, a + d, a + 2d, a + 3d,…

4, 10, 16, 22, …

Answer: 4, 10, 16, 22, …

Question 17

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

Given AP

3, 8, 13,…, 253
a = 3, d = 5, l = 253

nth terms of an AP

Let us find n for for last termΒ 

20th term from the last = (51 – 20 + 1)th term from start = 32nd term from start

Answer: 158

Question 18

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

Let the AP be:Β 

a, a + d, a + 2d,…

Sum of 4th and 8th terms = 24

Sum of 6th and 10th terms = 44

Subtract (1) from (2)

Substitute value of d in (1)

AP series:

a, a + d, a + 2d, …

-13, -8, -3, …

Answer: -13, -8, -3, …

Question 19

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

Salary in 1995 = 5000

Salary in 1996 = 5000 + 200 = 5200

Salary in 1997 = 5200 + 200 = 5400

and so on …

5000, 5200, 5400, … is an AP series

First term, a = 5200

Common difference, d = 200

Consider nth term will be 7000

n = 11 means in 11th year the salary will be 7000.

=> 1995 + 11 – 1 = 2005

In year 2005, the salary will be Rs. 7000.

Answer: 11th years, 2005

Why 2005 and not 2006? Watch the video.

Question 20

Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs.Β 1.75. If in the nth week, her weekly savings become Rs. 20.75, find n.

Savings in the 1st week = Rs. 5
Savings in the 2nd week = 5 + 1.75 = Rs. 6.75
Savings in the 3rd week = 6.75 + 1.75 = Rs. 8.5
Savings in the 4th week = 8.5 + 1.75 = Rs. 10.25
and so on…

5, 6.75, 8.5, 10.25, … is an AP series

First term, a = 5

Common difference, d = 1.75

Savings in the nth week are Rs. 20.75.

Answer: 10

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]]> Exercise 5.1 (NCERT) – Arithmetic Progressions, Class 10 – All Solutions https://mathemafia.com/exercise-5-1-ncert-arithmetic-progressions-class-10-all-solutions/ Wed, 07 May 2025 07:54:05 +0000 https://mathemafia.com/?p=18672 Exercise 5.1, chapter 5, Arithmetic Progressions, Class 10 - This exercise is primarily about identifying the AP series and finding few terms of a given AP series.

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Exercise 5.1 (NCERT) – Arithmetic Progressions, Class 10

Below are the Quick links for all questions of Exercise 5.1, Arithmetic Progressions, Class 10. Click a link to view the solution of corresponding question.

Chapter 5 Exercise 5.1 Class 10

Syal Mathematics

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Below are the solutions of every question of exercise 5.1Β of chapter 5, Arithmetic Progressions, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.Β Β 

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

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Fare for

1 km = β‚Ή15

2 km = 15 + 8 = β‚Ή23

3 km = 15 + 8 + 8 = β‚Ή31

4 km = 15 + 8 + 8 + 8 = β‚Ή39

and so on…

We get the list:

15,Β 23,Β 31,Β 39, …

This is an AP series because the difference between successive terms is constant = 8.

Answer: Yes, It is an AP series

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

Considering the original amount of air (a1)= 1 Unit

After the first removal, the remaining amount of air is:

After the second removal, the remaining amount of air is:

Let us find the common difference

Since aβ‚‚ βˆ’ a₁ β‰  a₃ βˆ’ aβ‚‚, the sequence is not an Arithmetic Progression (AP).

Answer: No, It is not an AP series

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

Digging cost

For 1 metre: β‚Ή150

For 2 metres: 150 + 50 = β‚Ή200

For 3 metres: 150 + 50 + 50 = β‚Ή250

For 4 metres: 150 + 50 + 50 + 50 = β‚Ή300

and so on…

We get the list:

150,Β 200,Β 250,Β 300, …

This is an AP series because the difference between successive terms is constant = 50.

Answer: Yes, It is an AP series

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

Amount in the account

For the 1st year:

For the 2nd year:

For the 3rd year:

We get the list:

10000, 10800, 11664, 12597, …

We can see that

A2 – A1 = 11664 – 10800 = 864

A3 – A2 = 12597 – 11664 = 933

Since aβ‚‚ βˆ’ a₁ β‰  a₃ βˆ’ aβ‚‚, the sequence is not an Arithmetic Progression (AP).

Answer: No, It is not an AP series

Question 2

Write first four terms of the AP, when the first term a and the common difference d are given as follows:

(i) a = 10, d = 10

First term
Common difference

First four terms are: a, a + d, a + 2d, a + 3d

a₁ = 10
aβ‚‚ = 10 + 10 = 20
a₃ = 10 + 20 = 30
aβ‚„ = 10 + 30 = 40

Answer: 10, 20, 30, 40

(ii) a = –2, d = 0

First term
Common difference

First four terms are: a, a + d, a + 2d, a + 3d

a₁ = –2
aβ‚‚ = –2 + 0 = –2
a₃ = –2 + 0 = –2
aβ‚„ = –2 + 0 = –2

Answer: -2, -2, -2, -2

(iii) a = 4, d = – 3

First term
Common difference

First four terms are: a, a + d, a + 2d, a + 3d

a₁ = 4
aβ‚‚ = 4 + (–3) = 1
a₃ = 1 + (–3) = –2
aβ‚„ = –2 + (–3) = –5

Answer: 4, 1, –2, –5

(iv) a = – 1, d = 1/2

First four terms are: a, a + d, a + 2d, a + 3d

Answer: –1, –1/2, 0, 1/2

(v) a = – 1.25, d = – 0.25

First term
Common difference

First four terms are: a, a + d, a + 2d, a + 3d

a₁ = –1.25
aβ‚‚ = –1.25 + (–0.25) = –1.50
a₃ = –1.50 + (–0.25) = –1.75
aβ‚„ = –1.75 + (–0.25) = –2.00

Answer: –1.25, –1.50, –1.75, –2.00

Question 3

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

(i) 3, 1, – 1, – 3, . . .

First term (a) = 3
Common difference (d) = aβ‚‚ – a₁ = 1 – 3 = –2

(ii) – 5, – 1, 3, 7, . . .

First term (a) = –5
Common difference (d) = aβ‚‚ – a₁ = (–1) – (–5) = –1 + 5 = 4

(iii) 1/3, 5/3, 9/3, 13/3

(iv) 0.6, 1.7, 2.8, 3.9, . . .

First term (a) = 0.6
Common difference (d) = aβ‚‚ – a₁ = 1.7 – 0.6 = 1.1

Question 4

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

(i) 2, 4, 8, 16, . . .

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Given sequence
2, 4, 8, 16, …

Checking the differences

aβ‚‚ βˆ’ a₁ = 4 βˆ’ 2 = 2
a₃ βˆ’ aβ‚‚ = 8 βˆ’ 4 = 4

Since aβ‚‚ βˆ’ a₁ β‰  a₃ βˆ’ aβ‚‚, the sequence is not an Arithmetic Progression (AP).

Answer: Not an AP

Important: If common difference for first three terms is not constant then we need not find any other term.

(iii) 2, 5/2, 3, 7/2, …

Given sequence

2, 2.5, 3, 3.5, …

Checking the differences

aβ‚‚ – a₁ = 2.5 – 2 = 0.5
a₃ – aβ‚‚ = 3 – 2.5 = 0.5
aβ‚„ – a₃ = 3.5 – 3 = 0.5

Since the differences are the same, the sequence is an Arithmetic Progression (AP).

Common difference (d) = 0.5

Three more terms

aβ‚… = 3.5 + 0.5 = 4.0
a₆ = 4.0 + 0.5 = 4.5
a₇ = 4.5 + 0.5 = 5.0

Answer: It is an AP series. Next three terms are 4, 4.5, 5

(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . .

Given sequence
–1.2, –3.2, –5.2, –7.2, …

Checking the differences

aβ‚‚ βˆ’ a₁ = –3.2 βˆ’ (–1.2) = –3.2 + 1.2 = –2
a₃ βˆ’ aβ‚‚ = –5.2 βˆ’ (–3.2) = –5.2 + 3.2 = –2
aβ‚„ βˆ’ a₃ = –7.2 βˆ’ (–5.2) = –7.2 + 5.2 = –2

Since the differences are the same, the sequence is an Arithmetic Progression (AP).

Common difference (d) = –2

Three more terms
aβ‚… = aβ‚„ + d = –7.2 + (–2) = –9.2
a₆ = aβ‚… + d = –9.2 + (–2) = –11.2
a₇ = a₆ + d = –11.2 + (–2) = –13.2

Answer: It is an AP series. Next three terms are –9.2, –11.2, –13.2

(iv) – 10, – 6, – 2, 2, . . .

Given sequence
–10, –6, –2, 2, …

Checking the differences

aβ‚‚ βˆ’ a₁ = –6 βˆ’ (–10) = –6 + 10 = 4
a₃ βˆ’ aβ‚‚ = –2 βˆ’ (–6) = –2 + 6 = 4
aβ‚„ βˆ’ a₃ = 2 βˆ’ (–2) = 2 + 2 = 4

Since the differences are the same, the sequence is an Arithmetic Progression (AP).

Common difference (d) = 4

Three more terms
aβ‚… = aβ‚„ + d = 2 + 4 = 6
a₆ = aβ‚… + d = 6 + 4 = 10
a₇ = a₆ + d = 10 + 4 = 14

Answer: It is an AP series. Next three terms are 6, 10, 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …

Given sequence
3, 3 + √2, 3 + 2√2, 3 + 3√2, …

Checking the differences
aβ‚‚ βˆ’ a₁ = (3 + √2) βˆ’ 3 = √2
a₃ βˆ’ aβ‚‚ = (3 + 2√2) βˆ’ (3 + √2) = 2√2 βˆ’ √2 = √2
aβ‚„ βˆ’ a₃ = (3 + 3√2) βˆ’ (3 + 2√2) = 3√2 βˆ’ 2√2 = √2

Since the differences are the same, the sequence is an Arithmetic Progression (AP).

Common difference (d) = √2

Three more terms

aβ‚… = aβ‚„ + d = (3 + 3√2) + √2 = 3 + 4√2
a₆ = aβ‚… + d = (3 + 4√2) + √2 = 3 + 5√2
a₇ = a₆ + d = (3 + 5√2) + √2 = 3 + 6√2

Answer: It is an AP series. Next three terms are 3 + 4√2, 3 + 5√2, 3 + 6√2

(vi) 0.2, 0.22, 0.222, 0.2222, . . .

Given sequence
0.2, 0.22, 0.222, 0.2222, …

Checking the differences
aβ‚‚ βˆ’ a₁ = 0.22 βˆ’ 0.2 = 0.02
a₃ βˆ’ aβ‚‚ = 0.222 βˆ’ 0.22 = 0.002

Since aβ‚‚ βˆ’ a₁ β‰  a₃ βˆ’ aβ‚‚, the sequence is not an Arithmetic Progression (AP).

Answer: Not an AP

(vii) 0, – 4, – 8, –12, . . .

Given sequence
0, βˆ’4, βˆ’8, βˆ’12, …

Checking the differences
aβ‚‚ βˆ’ a₁ = βˆ’4 βˆ’ 0 = βˆ’4
a₃ βˆ’ aβ‚‚ = βˆ’8 βˆ’ (βˆ’4) = βˆ’8 + 4 = βˆ’4
aβ‚„ βˆ’ a₃ = βˆ’12 βˆ’ (βˆ’8) = βˆ’12 + 8 = βˆ’4

Since the differences are the same, the sequence is an Arithmetic Progression (AP).

Common difference (d) = βˆ’4

Three more terms
aβ‚… = aβ‚„ + d = βˆ’12 + (βˆ’4) = βˆ’16
a₆ = aβ‚… + d = βˆ’16 + (βˆ’4) = βˆ’20
a₇ = a₆ + d = βˆ’20 + (βˆ’4) = βˆ’24

Answer: It is an AP series. Next three terms are βˆ’16, βˆ’20, βˆ’24

(viii) –1/2, –1/2, –1/2, –1/2, …

Given sequence
βˆ’1/2, βˆ’1/2, βˆ’1/2, βˆ’1/2, …

Checking the differences
aβ‚‚ βˆ’ a₁ = βˆ’1/2 βˆ’ (βˆ’1/2) = 0
a₃ βˆ’ aβ‚‚ = βˆ’1/2 βˆ’ (βˆ’1/2) = 0
aβ‚„ βˆ’ a₃ = βˆ’1/2 βˆ’ (βˆ’1/2) = 0

Since the differences are the same (all 0), the sequence is an Arithmetic Progression (AP).

Common difference (d) = 0

Three more terms
aβ‚… = aβ‚„ + d = βˆ’1/2 + 0 = βˆ’1/2
a₆ = aβ‚… + d = βˆ’1/2 + 0 = βˆ’1/2
a₇ = a₆ + d = βˆ’1/2 + 0 = βˆ’1/2

Answer: It is an AP series. Next three terms are βˆ’1/2, βˆ’1/2, βˆ’1/2

(ix) 1, 3, 9, 27, . . .

Given sequence
1, 3, 9, 27, …

Checking the differences

aβ‚‚ βˆ’ a₁ = 3 βˆ’ 1 = 2
a₃ βˆ’ aβ‚‚ = 9 βˆ’ 3 = 6

Since aβ‚‚ βˆ’ a₁ β‰  a₃ βˆ’ aβ‚‚, the sequence is not an Arithmetic Progression (AP).

Answer: Not an AP

(x) a, 2a, 3a, 4a, . . .

Given sequence
a, 2a, 3a, 4a, …

Checking the differences
aβ‚‚ βˆ’ a₁ = 2a βˆ’ a = a
a₃ βˆ’ aβ‚‚ = 3a βˆ’ 2a = a
aβ‚„ βˆ’ a₃ = 4a βˆ’ 3a = a

Since the differences are the same, the sequence is an Arithmetic Progression (AP).

Common difference (d) = a

Three more terms
aβ‚… = aβ‚„ + d = 4a + a = 5a
a₆ = aβ‚… + d = 5a + a = 6a
a₇ = a₆ + d = 6a + a = 7a

Answer: It is an AP series. Next three terms are 5a, 6a, 7a

(xi) a, a2, a3, a4, …

Given sequence
a, aΒ², aΒ³, a⁴, …

Checking the differences
aβ‚‚ βˆ’ a₁ = aΒ² βˆ’ a = a(a βˆ’ 1)
a₃ βˆ’ aβ‚‚ = aΒ³ βˆ’ aΒ² = aΒ²(a βˆ’ 1)

Since these differences depend on powers of a and are not constant (unless a = 1 or a = 0), the sequence is not an Arithmetic Progression (AP) in general. Try a = 2.

Since aβ‚‚ βˆ’ a₁ β‰  a₃ βˆ’ aβ‚‚, the sequence is not an Arithmetic Progression (AP).

Answer: Not an AP

(xii) √2, √8, √18, √32, . . .

Given sequence

√2, √8, √18, √32, . . .

which is same asΒ 
√2, 2√2, 3√2, 4√2, …

Checking the differences

aβ‚‚ βˆ’ a₁ = 2√2 βˆ’ √2 = √2
a₃ βˆ’ aβ‚‚ = 3√2 βˆ’ 2√2 = √2
aβ‚„ βˆ’ a₃ = 4√2 βˆ’ 3√2 = √2

Since the differences are the same, the sequence is an Arithmetic Progression (AP).

Common difference (d) = √2

Three more terms
aβ‚… = aβ‚„ + d = 4√2 + √2 = 5√2
a₆ = aβ‚… + d = 5√2 + √2 = 6√2
a₇ = a₆ + d = 6√2 + √2 = 7√2

Answer: It is an AP series. Next three terms are 5√2, 6√2, 7√2

(xiii) √3, √6, √9, √12, . . .

Given sequence
√3, √6, √9, √12, …

Checking the differences
aβ‚‚ βˆ’ a₁ = √6 βˆ’ √3 β‰ˆ 2.449 βˆ’ 1.732 = 0.717
a₃ βˆ’ aβ‚‚ = √9 βˆ’ √6 = 3 βˆ’ 2.449 = 0.551

Since aβ‚‚ βˆ’ a₁ β‰  a₃ βˆ’ aβ‚‚, the sequence is not an Arithmetic Progression (AP).

Answer: Not an AP

(xiv) 12, 32, 52, 72, . . .

Given sequence

12,32,52,72,…

Checking the differences

Since aβ‚‚ βˆ’ a₁ β‰  a₃ βˆ’ aβ‚‚, the sequence is not an Arithmetic Progression (AP).

Answer: Not an AP

(xv) 12, 52, 72, 73, . . .

Given sequence
1, 25, 49, 73, …

Checking the differences
aβ‚‚ βˆ’ a₁ = 25 βˆ’ 1 = 24
a₃ βˆ’ aβ‚‚ = 49 βˆ’ 25 = 24
aβ‚„ βˆ’ a₃ = 73 βˆ’ 49 = 24

Since the differences are the same, the sequence is an Arithmetic Progression (AP).

Common difference (d) = 24

Three more terms
aβ‚… = aβ‚„ + d = 73 + 24 = 97
a₆ = aβ‚… + d = 97 + 24 = 121
a₇ = a₆ + d = 121 + 24 = 145

Answer: It is an AP series. Next three terms are 97, 121, 145.

The post Exercise 5.1 (NCERT) – Arithmetic Progressions, Class 10 – All Solutions appeared first on Mathemafia.

]]> Class 10 – Chapter 5 – Arithmetic Progression – Last Minute Formula Revision https://mathemafia.com/class-10-chapter-5-arithmetic-progression-last-minute-formula-revision/ Thu, 06 Feb 2025 04:39:59 +0000 https://mathemafia.com/?p=9688 This short video covers all the formulas of chapter 5 (Arithmetic Progression, AP Series) of class 10. It is recommended to watch this before your exam to quickly recall all the formulas of this chapter.

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⏰ Last minute Formula Revision πŸ”₯ Arithmetic Progression, Chapter 5, Class 10

This short video covers all the formulas of chapter 5 (Arithmetic Progression, AP Series) of class 10. It is recommended to watch this before your exam to quickly recall all the formulas of this chapter.

The tutorial is available in Hinglish + 9 different languages:Β English, French, German, Hindi, Spanish, Indonesian, Italian, Japanese, Portuguese

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]]> Class 10 | Chapter 5 | All Solutions | Arithmetic Progression https://mathemafia.com/class-10-chapter-05-all-solutions-arithmetic-progression/ Fri, 17 Dec 2021 07:37:56 +0000 http://sh016.global.temp.domains/~mathemaf/?p=7360 This playlist covers all the questions, concepts and examples of chapter 5 (Arithmetic Progression) of class 10, NCERT book. Click a thumbnail to watch the tutorial. Watch complete chapter on YouTube

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This playlist covers all the questions, concepts and examples of chapter 5 (Arithmetic Progression) of class 10, NCERT book. Click a thumbnail to watch the tutorial.Β 

Watch complete chapter on YouTube

The post Class 10 | Chapter 5 | All Solutions | Arithmetic Progression appeared first on Mathemafia.

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