Class 10 Math Arithmetic Progression Most Important Questions with Solutions

Arithmetic progression, or AP, is a fundamental concept in mathematics, especially for students in class 10. It introduces the idea of sequences, patterns, and relationships between numbers, which is important for careers in finance, physics, and computer science. Acing the arithmetic progression. It not only strengthens your math basics but also enhances problem-solving skills that are important for higher studies. 

Here, we will delve into the detailed concepts of arithmetic progression, supplemented by important class 10 AP questions from our book and additional arithmetic progression class 10 questions, to ensure our students receive the best possible instruction.

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Chapter 5 Arithmetic Progressions: Important Questions

1. Given below are the details of an experiment using a bucket and a mug to understand water consumption. 

The bucket's volume is 30 litres and the mug's volume is 1/20 of the bucket. The bucket has 1 litre of water before the tap is turned on. The tap is filling the bucket at a constant rate of 0.1 litres per second. Every 30 seconds, he takes a mug full of water from the bucket.

i) Write an arithmetic progression for the volume of water in the bucket every 30 seconds.

ii) Find the volume of water in the bucket after exactly 7.5 minutes. Show your work.

(Note: Assume no spillage of water.)

Soln. 

In every 30 seconds:

• Water added by the tap: 

0.1 litres/second×30 seconds

=3 litres

• Water removed using the mug: 

1.5 litres.

The net gain in water every 30 seconds:

Net water added

=3−1.5

=1.5 litres

Let the initial volume of water in the bucket be a₁=1 litre

After 30 seconds(1st term): a₂ =1+1.5

=2.5 litres

After 60 seconds(2nd term): a₃ =2.5 +1.5

=4.0 litres

a₁=1 

Common difference d= 1.5

AP= 1,2.5,4.0,5.5,…

⇒7.5minutes=7.5×60

=450seconds.

Each interval is 30 seconds.

∴ n= 450/30

n= 15

a_n​=a_1​+(n−1)⋅d

a₁₅=1+(15−1).1.5

a₁₅=1+(14).1.5

a₁₅=1+21

a₁₅=22

i) Arithmetic progression for the volume of water in the bucket every 30 seconds is 

AP= 1,2.5,4.0,5.5,…

ii) The volume of water in the bucket after exactly 7.5 minutes is 22 litres

2. The ratio of the sum of the first 11 terms of an arithmetic progression to the sum of its first 21 terms is given by 1:4.

i) Show that 23a + 10d = 0, where a is the first term and d is the common difference of the arithmetic progression.

ii) Write an expression for the nth term of the above arithmetic progression only in terms of a and n.Show your work.

Soln. 

S₁₁/S₂₁= ¼

i) Show that 23a + 10d = 0,

S_n​=n/2​(2a+(n−1)d),

Using the given ratio

S₁₁​=11/2​(2a+(11−1)d),

S₁₁​=11/2​(2a+(10)d),…………………1

S₂₁​=21/2​(2a+(21−1)d),

S₂₁​=21/2​(2a+(20)d), …………………2

Put equation 1 and 2 in the given ratio

S₁₁/S₂₁= ¼

(11/2​(2a+10d))/(21/2​(2a+20d))= ¼

4 (11/2​(2a+10d)) = (21/2​(2a+20d))

44(2a+10d)=21(2a+20d).

88a+440d=42a+420d.

88a−42a+440d−420d=0.(Combine like terms)

46a+20d=0. (divide by 2)

23a+10d=0. 

Hence proved

ii) Write an expression for the nth term

23a+10d=0.

10d=−23a

d= (-23a/10)

T_n​=a+(n−1)d,

T_n​=a+(n−1)(-23a/10),

T_n​=a−(23a​/10)(n−1).

T_n​=a(1−(23/10)(n−1))

T_n​=a(10−(23n-23)/10

T_n​=a(10−23n+23)/10

T_n​=a(33−23n)/10

3. Answer the questions based on the given information.

Below is a house of cards, a structure created by stacking playing cards on top of each other in the shape of a pyramid. Each small triangle is made using 3 cards, and each layer has one less triangle than the layer below it.

Ankit and his friends were having a sleepover and wanted to do something fun. One of the friends suggested that they could make a house of cards.

i) Ankit and his friends want to use 3 cards in the top layer and 18 in the bottom layer.

Form an AP showing the number of cards in each layer starting from the top layer.

ii) Ankit is planning to make a pyramid with the top and bottom layer containing 15 and 138 cards respectively.

How many layers will such a pyramid have? Show your work.

iii) They have a total of 360 cards with them.

Find the maximum number of layers that Ankit and his friends can make using the cards they have, if they want to have 1 triangle (3 cards) at the top layer. Show your work.

Soln. 

i) Ankit and his friends want to use 3 cards in the top layer and 18 in the bottom layer.

Form an AP showing the number of cards in each layer starting from the top layer.

Top layer: 3 cards

Bottom layer: 18 cards

The number of cards in each layer forms an arithmetic progression (AP), where:

The first term a=3

The last term, l=18

The common difference d is to be determined.

The general formula for the n-th term of an AP is:

a_n​=a+(n−1)d

a_n=18

18=3+(n−1)d

(n−1)d=15

⟹d= 15/(n-1)

The AP showing the cards in each layer starting from the top layer is:

3,3+d,3+2d,…,18.

ii) Ankit is planning to make a pyramid with the top and bottom layer containing 15 and 138 cards respectively.

How many layers will such a pyramid have? Show your work.

a_n​=a+(n−1)d

Substitute the values a_n=138, a=15 , common difference is d=3(since each subsequent layer adds 3 cards

138=15+(n−1)(3)

138=15+3(n−1)

138=15+3n−3

138=3n+12

138−12=3n.

126=3n

n= 126/3

n=42

Number of layers in the pyramid is 42

iii) They have a total of 360 cards with them.

Find the maximum number of layers that Ankit and his friends can make using the cards they have, if they want to have 1 triangle (3 cards) at the top layer. Show your work.

The number of cards in the layers will be:

3,6,9,12,…

 Total number of cards:  S_n​=n/2 [2a+(n−1)d]

360=n/2 [2(3)+(n−1)(3)]

360=n/2 [6+3(n−1)].

360=n/2 [6+3n−3]

360=n/2 [3+3n].

720=n[3+3n]

720=3n+3n²

3n²+3n−720=0

n²+n−240=0

n= 15 or n = -16

The maximum number of layers that can be made is:15

4. The average of an Arithmetic Progression with 151 terms is zero. One of its terms is zero.

Which term of the Arithmetic Progression is zero? Show your steps.

Soln. 

Average= Sum of all terms / Number of terms

S₁₅₁​=0

S_n​= n/2 (a+l)

For n=151, S₁₅₁​=0

0 ​= 151/2 (a+l)

This implies:

a+l=0⇒l=−a

Thus, the first term, a and the last term, l are negatives.

The AP has 151 terms, which is an odd number. In an AP with an odd number of terms, the middle term is the average of the AP. Since the average is 0, the middle term must be 0.

Middle term=(151+1)/2

Middle term= 76

The 76th term of the AP is 0.

5. Parth was receiving spam calls from a telemarketing centre. He got the first call at 2:48 pm as shown below.

If the telemarketer continues to call using the same pattern, at what time will Parth receive the 13th call? Show your work.

Soln. 

From the missed call times:

• 2:48 PM → 2:52 PM → 2:58 PM → 3:06 PM → 3:16 PM.

Let’s calculate the time differences between consecutive calls:

1. From 2:48 PM to 2:52 PM → 4 minutes
2. From 2:52 PM to 2:58 PM → 6 minutes
3. From 2:58 PM to 3:06 PM → 8 minutes
4. From 3:06 PM to 3:16 PM → 10 minutes

Thus, the intervals between calls are increasing by 2 minutes2 \, \text{minutes}2minutes each time. This forms an arithmetic progression (AP) for the time intervals..

The sequence of time intervals is:

4,6,8,10,……

Here:

• The first term (a) = 4 minutes
• The common difference (d) = 2 minutes

S_n​=n/2​[2a+(n−1)d],

Where n=12 (number of time intervals),

a=4 (first term),

d=2 (common difference).

S₁₂=12/2​[2(4)+(12−1)(2)],

S₁₂=6[8+11(2)]

S₁₂=6[8+22]

S₁₂=6×30

=180minutes

2:48PM+3hours=5:48PM

Parth will receive the 13th call at:5:48PM

7. Shivam bought a large quantity of marigold flowers to decorate his house for a family function. He used 630 flowers to recreate the pattern shown below. He used 7 flowers in the shortest garland and 35 flowers in the longest garland.

(Note: The figure is for visual representation only.)

If he kept the difference between two consecutive garlands the same, how many garlands did he make? Show your steps.

Soln. 

• First term a=7 (shortest garland),
• Last term l=35 (longest garland),
• Total sum S_n=630
• n is the total number of garlands (unknown),
• Common difference d (unknown).

The sum of the first n terms of an AP is given by:

S_n​=n/2(a+l)

630=n/2​(7+35)

630=n/2​(42)

1260=n⋅42.

n=1260​/42

n=30

The number of garlands Shivam made is:30

7. √2, √18, √50, √98...

Is the above pattern in AP? Justify your answer.

Soln. 

d=a₂​−a₁​=a₃​−a_2​=a_4​−a₃​(constant)

First term: a₁=√2 ≈ 1.141

Second term:a₂=√18 

=√9.2

=3√2 ≈ 1.141

Third term:a₃=√50

=√25.2

=5√92 ≈ 7.071

Fourth term:a₄=√98

=√49.2

=7√2 ≈9.899

a₂​−a_1​=4.243−1.414

=2.829,

a₃−a₂=7.071−4.243

=2.828

a₄−a₃=9.899−7.071

=2.828

The differences are approximately:

2.829,2.828,2.828

While they are very close, they are not exactly the same due to rounding differences. However, notice that each term follows the pattern:

a_n​=(2n−1) √2

d=a_n+1​−a_n

a_n+1​=(2(n+1)−1)√2 and a_n=(2n−1)2a_n = (2n-1)√2:

d=[(2(n+1)−1)√2 ​]−[(2n−1)√2​].

=√2

Yes, the sequence is AP

8. Sana decided to start practicing for an upcoming marathon. She decided to gradually increase the duration. She ran for 10 mins on day 1 and increased the duration by 5 minutes every day.

From which day onwards will she be running for 2.5 hours or more? Show your work.

Soln. 

On Day 1, she ran for 10 minutes10

• First term (a) = 10 minutes
• Common difference (d) = 5 minutes,
• We need to find the day (n) when the duration is 2.5 hours or more.

Convert 2.5 hours into minutes:

2.5 hours=2.5×60

=150 minutes

a_n​=a+(n−1)d,

150=10+(n−1)5.

140=5(n−1)

140/5=(n−1)

n−1=28

n= 29

From Day 29 onwards, Sana will be running for 2.5 hours (150 minutes) or more

9. Kevin is baking a tall layered wedding cake as shown below. The customer has ordered a 111 kg cake and 12 layers.

(Note: The image is for visual representation only.)

For the cake to stand properly, he makes the bottom-most cake of 17.5 kg and reduced the weight of each layer such that the difference in the weights of the consecutive layers is the SAME.

i) By what weight does he reduce each subsequent layer?

ii) What is the weight of the lightest cake layer?

Show your work.

Soln. 

i) By what weight does he reduce each subsequent layer?

• The total weight of the cake = 111 kg,
• The number of layers = 12,
• The bottom-most layer weighs 17.5 kg,
• The weight of each layer decreases by a constant amount.

Let the weight of each layer form an arithmetic progression (AP) where:

• The first term a=17.5 (weight of the bottom-most layer),
• The number of terms n=12
• The sum of all terms S_n=111

S_n​=n/2​[2a+(n−1)d],

111= 12/2 [2(17.5)+(12−1)d]

111=6[2(17.5)+11d].

111=6[35+11d]

(111/6)=35+11d.

18.5=35+11d.

11d=18.5−35.

11d=−16.5

d= −16.5/11

d= −1.5.

ii) What is the weight of the lightest cake layer?

a_n​=a+(n−1)d,

a_12​=17.5+(12−1)(−1.5).

a_12​=17.5+11(−1.5).

a_12​=17.5−16.5

a_12​=1.0.

Understanding Arithmetic Progression (Class 10) 

Arithmetic progression is a sequence of numbers where the difference between any two consecutive terms is always constant. We refer to this difference as a Common Difference. For instance, in the sequence 2, 5, 8, 11,..., there is a difference of 3 between each pair of consecutive terms, indicating an arithmetic progression.

The general form of an arithmetic progression is written as: a, a + d, a + 2d, a + 3d,... where:

• A is the first term.
• D is the common difference, and
• The terms increase (or decrease) by adding D to the previous term.

Key Terms in Arithmetic Progression for Class 10

To excel in AP, focus on the following key topics, which are often covered in important questions in arithmetic progression class 10 with solutions pdf and similar resources:

• Finding the nth term: Given the first term and the common difference, how do you find the nth term of the sequence? This is one of the most fundamental questions in AP.
• Sum of First n Terms: Knowing how to calculate the sum of the first n terms of an AP is crucial, especially in word problems.
• Word Problems: AP often appears in real-world situations. Be prepared to translate a real-life scenario into an arithmetic progression problem and solve it.
• Common Difference: Identifying the common difference in a sequence and using it to solve problems is important.

Benefits of Solving Class 10 Arithmetic Progression Extra Questions

I understand that you are already aware of the benefits of using the class 10 AP extra questions, but let me reiterate the importance of using these questions:

• Solving extra AP questions in class 10 helps improve analytical and logical thinking.  These questions aim to enhance your reasoning abilities, equipping you to tackle not only AP issues but also intricate mathematical problems.
• Regular practice of arithmetic progression class 10 important questions ensures that you're well-prepared for your board exams. These questions cover a range of difficulty levels, allowing you to confidently approach any problem that may come up in the exam.
• By solving the important question of AP class 10, you’ll learn to manage time more effectively during exams. The more you practice, the quicker you become at solving complex questions, allowing you to complete your exam on time without feeling rushed.

• As arithmetic progression is an important chapter in Class 10, regularly practicing extra questions ensures that you perform well in exams, thus boosting your overall score in mathematics.
• Solving complex problems related to AP helps in building your confidence. When you know that you've worked through the toughest problems, you can approach the exam with a calm and composed mindset.

How to Use Class 10 Arithmetic Progression Questions

When studying arithmetic progression, working through Class 10 AP extra questions is one of the most effective ways to solidify your understanding of the topic. Here’s how you can use them:

• Before diving into the extra questions, ensure that you have a clear understanding of the basic concepts of AP. Go through your textbooks and revise the formulas, especially the general formula for the nth term of an AP and the sum of the first n terms.
• Start by solving basic AP questions, where you are required to find the nth term or the common difference. Once you’re comfortable with these, move on to more tough problems.
• When you go through the extra questions in class 10 maths chapter 5, pay special attention to the problems that resemble previous exam papers. Solving these types of questions helps you familiarize yourself with the format and difficulty level of the actual exam.
• As you practice, identify which types of questions are more difficult for you, whether it's calculating the sum of an AP, finding a specific term, or solving word problems. Concentrate on these areas by revisiting similar additional questions and reviewing their solutions.

Arithmetic Progression Class 10 Important Questions makes a strong base for future math and problem-solving. Whether it's understanding the basic concepts or handling more advanced problems, practicing with arithmetic progression (10 extra questions) and important questions is essential.  These resources enhance your analytical skills, improve your exam performance, and prepare you for more complex topics in mathematics. By using the extra questions and solutions, either from PDFs or practice books, and regularly reviewing your progress, you’ll ace your Class 10 exams.