Which of the following is NOT an example of a situation that can be modelled by an AP?

Amount becoming 5/4 times of itself after every 3 years.

Salaries increasing by a fixed amount annually.

Lengths of ladder rungs decreasing uniformly.

Amount becoming 5/4 times of itself after every 3 years.

The number of seats in rows of an auditorium increasing by a fixed number.

Assertion (A): The sequence 2, 2, 2, 2, ... is an Arithmetic Progression. Reason (R): The common difference of an AP can be zero.

A and R both correct, R explains A

Assertion (A): If the nth term of an AP is 3n + 1, then its common difference is 3. Reason (R): The common difference of an AP is the coefficient of 'n' in the expression for the nth term (if it's linear in 'n').

A and R both correct, R explains A

Is this CBSE Class 10 Maths Chapter 5 quiz free?

Yes, absolutely! This quiz on Arithmetic Progressions is completely free for all students. No hidden costs or subscriptions.

How many questions are in the quiz?

The main 'Game' mode of this quiz features 40 multiple-choice questions, carefully designed to cover all aspects of the chapter.

Is this quiz aligned with the NCERT syllabus?

Yes, all questions in this quiz are strictly based on the NCERT Class 10 Maths textbook and the latest CBSE syllabus for Chapter 5, Arithmetic Progressions.

What topics does this quiz cover?

This quiz covers the definition of an AP, common difference, finding the nth term of an AP, and calculating the sum of the first n terms of an AP.

Can I re-attempt the quiz?

Yes, you can attempt the quiz as many times as you like to practice and improve your understanding of Arithmetic Progressions.

Are there explanations for the answers?

Yes, detailed explanations are provided for every question, helping you understand the concepts and correct any mistakes.

Home › CBSE › Class 10 › Maths › ARITHMETIC PROGRESSIONS

CBSE · Class 10 · 🧮 Maths · Chapter 5

ARITHMETIC PROGRESSIONS

Arithmetic Progression (AP)Common difference (d)nth term of an APSum of first n terms of an APGeneral form of an AP

Chapter 5, 'Arithmetic Progressions', introduces students to a fundamental concept in mathematics: sequences where the difference between consecutive terms is constant. You will learn to identify an AP, find its common difference, calculate the nth term using the formula a_n = a + (n-1)d, and determine the sum of the first n terms using S_n = n/2 [2a + (n-1)d] or S_n = n/2 (a + l). Understanding APs is crucial for further studies in sequences and series and has applications in various real-life scenarios.

Introduction to AP and its basic concepts

AP kya hai? (What is an AP?)

Arithmetic Progression (AP) ek sequence of numbers hai jismein har term (first term ko chhodkar) apne preceding term mein ek fixed number add karke milta hai.
Ye fixed number ko common difference (d) kehte hain.
Common difference (d) positive, negative ya zero ho sakta hai.

AP ko kaise pehchanein? (How to identify an AP?)

Kisi bhi sequence ko AP kehne ke liye, consecutive terms ke beech ka difference constant hona chahiye.
Check karne ke liye: a₂ - a₁, a₃ - a₂, a₄ - a₃, ... calculate karo. Agar ye sab same hain, toh sequence AP hai.

Example:

2, 4, 6, 8, ...

4 - 2 = 2
6 - 4 = 2
8 - 6 = 2
Common difference d = 2. Ye ek AP hai.

100, 70, 40, 10, ...

70 - 100 = -30
40 - 70 = -30
10 - 40 = -30
Common difference d = -30. Ye bhi ek AP hai.

3, 3, 3, 3, ...

3 - 3 = 0
3 - 3 = 0
Common difference d = 0. Ye bhi ek AP hai.

1, 4, 9, 16, ...

4 - 1 = 3
9 - 4 = 5
16 - 9 = 7
Differences same nahi hain. Ye AP nahi hai.

Key Points:

First term ko a ya a₁ se denote karte hain.
Common difference ko d se denote karte hain.
d = a_{k+1} - a_k (kisi bhi term mein se uske pehle wala term minus karke).

📖Definition

Arithmetic Progression (AP): A sequence of numbers jismein har term (first term ko chhodkar) apne preceding term mein ek fixed number add karke milta hai. Ye fixed number common difference (d) hota hai.

❗Important

Common difference (d) positive, negative ya zero ho sakta hai. Isko calculate karne ke liye d = a₂ - a₁ ya d = a₃ - a₂ use karte hain.

General form of an AP

AP ka General Representation

Agar first term a hai aur common difference d hai, toh AP ko aise represent karte hain:

a, a + d, a + 2d, a + 3d, ...

Har term ko uske position ke hisaab se likhte hain:
First term (a₁) = a
Second term (a₂) = a + d
Third term (a₃) = a + 2d
Fourth term (a₄) = a + 3d
Aur isi tarah aage bhi.

Example:

Agar a = 5 aur d = 3 hai, toh AP kya hogi?
a₁ = 5
a₂ = 5 + 3 = 8
a₃ = 5 + 2(3) = 5 + 6 = 11
a₄ = 5 + 3(3) = 5 + 9 = 14
AP: 5, 8, 11, 14, ...

Agar a = 10 aur d = -2 hai, toh AP kya hogi?
a₁ = 10
a₂ = 10 + (-2) = 8
a₃ = 10 + 2(-2) = 10 - 4 = 6
a₄ = 10 + 3(-2) = 10 - 6 = 4
AP: 10, 8, 6, 4, ...

🧮Formula

General Form of AP: a, a + d, a + 2d, a + 3d, ... Jahan a first term hai aur d common difference hai.

nth Term of an AP

nth Term ka Formula (Formula for nth Term)

Kisi AP ka nth term (a_n) find karne ke liye formula hai:

a_n = a + (n - 1)d Jahan:

a_n = nth term
a = first term
n = number of terms (ya term ki position)
d = common difference

Derivation of nth Term Formula

a₁ = a
a₂ = a + d = a + (2 - 1)d
a₃ = a + 2d = a + (3 - 1)d
a₄ = a + 3d = a + (4 - 1)d
Pattern observe karo: jo term number hai, d ke coefficient mein usse ek kam hai.
Isliye, a_n = a + (n - 1)d.

Important Points:

Agar AP mein m terms hain, toh last term l ko a_m se denote karte hain, aur l = a + (m - 1)d.
Agar aapko end se nth term nikalna hai, toh formula hai:

a_n (from end) = l - (n - 1)d Jahan l last term hai. Alternative method: Pehle total number of terms m find karo. Phir end se nth term, start se (m - n + 1)th term hota hai. a_{m-n+1} = a + (m - n + 1 - 1)d = a + (m - n)d.

Example:

AP: 2, 7, 12, ... ka 10th term find karo.
a = 2
d = 7 - 2 = 5
n = 10
a_{10} = a + (10 - 1)d = 2 + 9(5) = 2 + 45 = 47

AP: 21, 18, 15, ... ka kaun sa term -81 hai?
a = 21
d = 18 - 21 = -3
a_n = -81
a_n = a + (n - 1)d
-81 = 21 + (n - 1)(-3)
-81 - 21 = -3(n - 1)
-102 = -3(n - 1)
34 = n - 1
n = 35
Toh, 35th term -81 hai.

🧮Formula

nth Term of an AP: a_n = a + (n - 1)d Jahan a_n nth term, a first term, n term number, d common difference.

💡Tip

Agar question mein 'kaun sa term' pucha hai, toh n find karna hai. Agar nth term ki value puchi hai, toh a_n find karna hai.

Sum of First n Terms of an AP

Sum of First n Terms ka Formula (Formula for Sum of First n Terms)

Kisi AP ke first n terms ka sum (S_n) find karne ke liye do main formulas hain:

S_n = \frac{n}{2} [2a + (n - 1)d]
S_n = \frac{n}{2} [a + l] (Jab last term l pata ho)

Jahan:

S_n = sum of first n terms
a = first term
n = number of terms
d = common difference
l = last term (a_n ke barabar)

Derivation of Sum Formula

Let S_n = a + (a+d) + (a+2d) + ... + (a+(n-1)d) --- (1) Terms ko reverse order mein likho: S_n = (a+(n-1)d) + (a+(n-2)d) + ... + a --- (2)

(1) aur (2) ko add karne par: 2S_n = [a + (a+(n-1)d)] + [(a+d) + (a+(n-2)d)] + ... + [(a+(n-1)d) + a] 2S_n = [2a + (n-1)d] + [2a + (n-1)d] + ... + [2a + (n-1)d] (n times) 2S_n = n [2a + (n-1)d] S_n = \frac{n}{2} [2a + (n-1)d]

Since l = a + (n-1)d, hum isko substitute kar sakte hain: S_n = \frac{n}{2} [a + a + (n-1)d] S_n = \frac{n}{2} [a + l]

Important Points:

Agar S_n diya ho aur a_n nikalna ho, toh:

a_n = S_n - S_{n-1} (Ye formula tab use hota hai jab terms ka sum diya ho aur koi particular term find karna ho).

S_1 = a_1 = a

Example:

AP: 2, 7, 12, ... ke first 10 terms ka sum find karo.
a = 2
d = 5
n = 10
S_{10} = \frac{10}{2} [2(2) + (10 - 1)5]
S_{10} = 5 [4 + 9(5)]
S_{10} = 5 [4 + 45]
S_{10} = 5 [49] = 245

Agar kisi AP ka S_n = 3n^2 + 5n hai, toh uska 2nd term (a₂) find karo.
S_1 = 3(1)^2 + 5(1) = 3 + 5 = 8
S_2 = 3(2)^2 + 5(2) = 3(4) + 10 = 12 + 10 = 22
a_1 = S_1 = 8
a_2 = S_2 - S_1 = 22 - 8 = 14

🧮Formula

Sum of First n Terms of an AP:

S_n = \frac{n}{2} [2a + (n - 1)d]
S_n = \frac{n}{2} [a + l] (Jab last term l pata ho)

Relation between S_n and a_n: a_n = S_n - S_{n-1}

🚧Misconception

Students often confuse n (number of terms) with a_n (the nth term's value). Dhyan rakho ki n ek position hai aur a_n us position par value.

Properties and Applications of AP

AP ki Properties (Properties of AP)

Agar kisi AP ke har term mein same constant number add ya subtract kiya jaye, toh resultant sequence bhi AP hota hai, aur common difference same rehta hai.

Example: AP 2, 4, 6, 8 (d=2). Har term mein 3 add kiya: 5, 7, 9, 11 (d=2).

Agar kisi AP ke har term ko same non-zero constant se multiply ya divide kiya jaye, toh resultant sequence bhi AP hota hai, aur common difference bhi multiply/divide ho jata hai us constant se.

Example: AP 2, 4, 6, 8 (d=2). Har term ko 2 se multiply kiya: 4, 8, 12, 16 (d=4).

Agar a, b, c AP mein hain, toh 2b = a + c hota hai. b ko Arithmetic Mean kehte hain a aur c ka.

b - a = c - b
2b = a + c
b = \frac{a+c}{2}

AP ke Applications (Applications of AP)

Daily life scenarios: Salary increments, ladder rungs, fixed deposits with simple interest, etc.
Problem Solving:
Kisi series mein terms count karna (e.g., 2-digit numbers divisible by 3).
Total amount ya distance calculate karna (e.g., total distance covered in a race, total money saved).

Example Problems:

Problem: 10 aur 250 ke beech kitne multiples of 4 hain?
First multiple of 4 after 10 is 12 (a = 12).
Last multiple of 4 before 250 is 248 (a_n = 248).
Common difference d = 4.
a_n = a + (n - 1)d
248 = 12 + (n - 1)4
236 = (n - 1)4
59 = n - 1
n = 60
Toh, 10 aur 250 ke beech 60 multiples of 4 hain.

Problem: Ek construction company ko ek project mein delay ke liye penalty deni padti hai. First day ke liye ₹200, second day ke liye ₹250, third day ke liye ₹300, aur aise hi har succeeding day ke liye ₹50 zyada. Agar contractor ne 30 din ka delay kiya, toh usko total kitni penalty deni padegi?
Ye ek AP hai jismein:
a = 200
d = 50
n = 30
Total penalty S_{30} hogi:
S_n = \frac{n}{2} [2a + (n - 1)d]
S_{30} = \frac{30}{2} [2(200) + (30 - 1)50]
S_{30} = 15 [400 + 29(50)]
S_{30} = 15 [400 + 1450]
S_{30} = 15 [1850]
S_{30} = 27750
Contractor ko total ₹27,750 ki penalty deni padegi.

⭐Remember

Agar a, b, c AP mein hain, toh b ko a aur c ka Arithmetic Mean kehte hain, aur b = \frac{a+c}{2}.

💡Tip

Word problems mein a, d, n, a_n, ya S_n ko correctly identify karna sabse important step hai. Question ko dhyan se padho!

Easy (15)

An arithmetic progression is a list of numbers where each term is obtained by adding a fixed number to the preceding term, except the first term. What is this fixed number called?
- Common factor
- Common difference (correct)
- Common ratio
- Common multiple
Why: The fixed number added in an AP is called the common difference.
Which of the following sequences is an Arithmetic Progression?
- 2, 4, 8, 16, ...
- 1, 3, 5, 7, ... (correct)
- 1, 1, 2, 3, 5, ...
- 1, 1/2, 1/3, 1/4, ...
Why: 1, 3, 5, 7, ... has a constant common difference of 2.
What is the common difference of the AP: 10, 7, 4, 1, ...?
- 3
- -3 (correct)
- 10
- 7
Why: The common difference is found by subtracting any term from its succeeding term.
The formula for the nth term of an AP is a_n = a + (n - 1)d. What does 'a' represent in this formula?
- The last term
- The common difference
- The first term (correct)
- The number of terms
Why: 'a' represents the first term of the arithmetic progression.
If the first term of an AP is 5 and the common difference is 3, what is the second term?
- 5
- 8 (correct)
- 2
- 11
Why: The second term is the first term plus the common difference.
Which of the following is NOT an example of a situation that can be modelled by an AP?
- Salaries increasing by a fixed amount annually.
- Lengths of ladder rungs decreasing uniformly.
- Amount becoming 5/4 times of itself after every 3 years. (correct)
- The number of seats in rows of an auditorium increasing by a fixed number.
Why: An amount increasing by a fixed multiple is geometric, not arithmetic.
If the terms of an AP are decreasing, what can be said about its common difference?
- It is positive.
- It is negative. (correct)
- It is zero.
- It can be any real number.
Why: A decreasing AP has a negative common difference.
What is the general form of an Arithmetic Progression?
- a, ar, ar², ar³, ...
- a, a+d, a+2d, a+3d, ... (correct)
- a, a/d, a/d², a/d³, ...
- a, a², a³, a⁴, ...
Why: The general form of an AP starts with 'a' and adds 'd' to each subsequent term.
The common difference of an AP can be positive, negative, or zero.
Why: The common difference can indeed be positive, negative, or zero.
If all terms in a sequence are the same, it is still considered an Arithmetic Progression.
Why: A sequence with all identical terms has a common difference of zero, making it an AP.
The sum of the first n terms of an AP can be found using the formula S = n/2 [2a + (n – 1) d].
Why: This is the correct formula for the sum of the first n terms of an AP.
The fixed number added to the preceding term in an AP is called the _________ difference.
Why: The fixed number is known as the common difference.
The nth term of an AP is given by a_n = a + (n – 1) _________.
Why: The formula for the nth term includes the common difference 'd'.
If the first term of an AP is 'a' and the common difference is 'd', the third term is a + _________d.
Why: The third term is a + 2d, following the pattern a + (n-1)d.
The sum of the first n terms of an AP, when the last term 'l' is known, is S = n/2 (a + _________).
Why: The formula for sum with last term 'l' is S = n/2 (a + l).

Medium (15)

Find the 10th term of the AP: 2, 7, 12, ...
- 47 (correct)
- 52
- 37
- 42
Why: Using a_n = a + (n-1)d, the 10th term is 47.
Which term of the AP: 21, 18, 15, ... is -81?
- 34th
- 35th (correct)
- 36th
- 37th
Why: Solving a_n = a + (n-1)d for n gives the term number.
Find the sum of the first 20 terms of the AP: 1, 3, 5, 7, ...
- 380
- 400 (correct)
- 420
- 440
Why: Using S_n = n/2 [2a + (n-1)d], the sum is 400.
If the 3rd term of an AP is 7 and the 7th term is 15, find the common difference.
- 1
- 2 (correct)
- 3
- 4
Why: Set up two equations using a_n = a + (n-1)d and solve for d.
How many two-digit numbers are divisible by 3?
- 29
- 30 (correct)
- 31
- 32
Why: The two-digit numbers divisible by 3 form an AP; find 'n'.
Match the following AP terms with their descriptions:
Why: This matches the standard notations used in Arithmetic Progressions.
Match the given sequences with their common differences:
Why: Calculate d = a₂ - a₁ for each sequence.
Match the APs with their first terms:
Why: The first term is simply the initial term of the sequence.
What is the common difference if the 5th term of an AP is 19 and the 8th term is 31?
Why: The common difference is (31-19)/(8-5) = 12/3 = 4.
If the first term of an AP is 10 and the common difference is -2, what is the 6th term?
Why: Using a_n = a + (n-1)d, a₆ = 10 + (6-1)(-2) = 10 - 10 = 0.
Arrange the steps to find the sum of the first 100 natural numbers using the AP formula:
Why: The steps involve identifying parameters, applying the formula, substituting, and calculating.
Arrange the steps to determine if a given list of numbers forms an AP:
Why: An AP is identified by a constant difference between consecutive terms.
Observe the pattern of squares and identify the next figure in the sequence if it represents an AP of increasing side lengths.
- Square with side 5 units (correct)
- Square with side 6 units
- Square with side 4 units
- Square with side 7 units
Why: The sequence of side lengths is 2, 3, 4, so the next is 5.
A ladder has rungs that decrease uniformly in length. If the bottom rung is 45 cm and the top rung is 31 cm, and there are 8 rungs in total, what is the length of the 4th rung from the bottom?
- 39 cm (correct)
- 41 cm
- 37 cm
- 43 cm
Why: The AP is 45, 43, 41, 39, ..., so the 4th rung is 39 cm.
Identify the common difference 'd' from the given AP representation on a number line.
- 2
- 3 (correct)
- 4
- 1
Why: The points are at 1, 4, 7, 10, so the common difference is 3.

Hard (10)

The sum of the first 15 terms of an AP is 105, and its first term is 3. What is the common difference?
Why: Using the sum formula, d = 0.4.
How many terms are there in the AP: 7, 13, 19, ..., 205?
Why: Solving for 'n' in the nth term formula gives 34 terms.
If the sum of the first 'n' terms of an AP is given by S_n = 3n² + 5n, what is the common difference of the AP?
Why: a₁ = S₁ = 8, a₂ = S₂ - S₁ = 22 - 8 = 14. So d = 14 - 8 = 6.
The 17th term of an AP exceeds its 10th term by 7. What is the common difference of the AP?
Why: a₁₇ - a₁₀ = (a+16d) - (a+9d) = 7d. Given this is 7, so 7d=7, d=1.
Assertion (A): The sequence 2, 2, 2, 2, ... is an Arithmetic Progression. Reason (R): The common difference of an AP can be zero.
Why: A sequence with a common difference of zero is an AP, so both are correct and R explains A.
Assertion (A): If the nth term of an AP is 3n + 1, then its common difference is 3. Reason (R): The common difference of an AP is the coefficient of 'n' in the expression for the nth term (if it's linear in 'n').
Why: If a_n = An + B, then d = A. Both A and R are correct and R explains A.
A man saves ₹1000 in the first month, ₹1200 in the second month, ₹1400 in the third month, and so on. If he continues this pattern, what will be his total savings after 12 months?
- ₹25200 (correct)
- ₹26400
- ₹24000
- ₹27600
Why: The savings form an AP with a=1000, d=200. S₁₂ = 12/2 [2(1000) + 11(200)] = 6 [2000 + 2200] = 6 * 4200 = 25200.
A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, what is the value of the first prize?
- ₹160 (correct)
- ₹140
- ₹180
- ₹120
Why: S₇ = 700, n=7, d=-20. 700 = 7/2 [2a + 6(-20)]. Solve for a.
A spiral is made up of successive semicircles with centres alternately at A and B, starting with centre A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ... as shown in the figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?
- 143 cm (correct)
- 182 cm
- 132 cm
- 154 cm
Why: The lengths are π(0.5), π(1.0), π(1.5), ... which is an AP with a=0.5π, d=0.5π. S₁₃ = 13/2 [2(0.5π) + 12(0.5π)] = 13/2 [π + 6π] = 13/2 [7π] = 91π/2. Using π=22/7, S₁₃ = 91/2 * 22/7 = 13 * 11 = 143 cm.
A tree plantation drive is organized where students of Class I to XII are involved. Each class plants trees equal to its class number. For example, Class I plants 1 tree, Class II plants 2 trees, and so on, up to Class XII. If there are two sections for each class, how many trees are planted in total?
- 156 (correct)
- 78
- 132
- 144
Why: Trees per section: 1+2+...+12 = S₁₂ = 12/2 (1+12) = 6 * 13 = 78. For two sections, total = 78 * 2 = 156.