Important Questions for Class 10 Maths Chapter 5: Arithmetic Progression

Important questions for Class 10 Maths Chapter 5 on Arithmetic Progression are available here to help students get ready for the 2024-2025 board exams. These questions follow the NCERT book and CBSE syllabus. We’ve made these questions based on the exam style, past papers, exam trends, and the latest 2024-25 sample papers. Solving these questions can help students get high scores in their Maths exams. They can also check their answers using the provided solutions. So, it’s a good idea for students to practice these questions well. This will boost their confidence and improve their skills.

Important Questions For Class 10 Chapter 5 With Solutions

Question: Define Arithmetic Progression (AP).

Answer: Arithmetic Progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant.

Question: What is the common difference in an AP?

Answer: The common difference in an AP is the constant value by which each term differs from its preceding term.

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

Answer: Given AP is: 3, 15, 27, 39, …
First term, a = 3
Common difference, d = a₂ − a₁ = 15 − 3 = 12

We know that,
a_n = a + (n − 1) d

Therefore,

a₅₄ = a + (54 − 1) d
= 3 + (53) (12)
= 3 + 636
a₅₄ = 639

We have to find the term of this AP.which is 132 more than a₅₄, i.e. 771.
Let nth term be 771.
a_n = a + (n − 1) d
771 = 3 + (n − 1) 12
768 = (n − 1) 12
⇒ (n − 1) = 64
⇒ n = 65

Therefore, the 65th term is 132 more than the 54th term of the given AP.

Alternate Method:

Let nth term be 132 more than 54th term.
n = 54 + (132/12)

= 54 + 11

= 65th term

Also Check: Important Questions for Class 10 Maths Chapter 1 – Real Numbers

Question: If the first term of an AP is ‘a’ and the common difference is ‘d’, what is the nth term of the AP?

Answer: The nth term

${𝑇}_{𝑛}$

Tn​ of an AP is given by the formula

${𝑇}_{𝑛}=𝑎+(𝑛-1)𝑑$

Tn​=a+(n−1)d.

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

Answer: Given AP: 11, 8, 5, 2, …

First term, a = 11

Common difference, d = a₂ − a₁ = 8 − 11 = −3

Let −150 be the nth term of this AP.

As we know, for an AP,

a_n = a + (n − 1) d

-150 = 11 + (n – 1)(-3)

-150 = 11 – 3n + 3

⇒ -164 = -3n

⇒ n = 164/3

Clearly, n is not an integer but a fraction.

Therefore, – 150 is not a term of the given AP.

Question: How can you find the number of terms in an AP when given the first term, last term, and common difference?

Answer: The number of terms in an AP can be found using the formula

$𝑛=\frac{𝑙-𝑎}{𝑑}+1$

n=dl−a​+1, where

$𝑎$

a is the first term,

$𝑙$

l is the last term, and

$𝑑$

d is the common difference.

Question: What is the sum of the first ‘n’ terms of an AP?

Answer: The sum

${𝑆}_{𝑛}$

Sn​ of the first ‘n’ terms of an AP is given by the formula

${𝑆}_{𝑛}=\frac{𝑛}{2}(2𝑎+(𝑛-1)𝑑)$

Sn​=2n​(2a+(n−1)d).

Also Check: Important Questions for Class 10 Maths Chapter 4 Quadratic Equations

Question: If the 5th term of an AP is 17 and the 10th term is 32, find the common difference.

Answer:

$𝑑=\frac{{𝑇}_{10}-{𝑇}_5}{10-5}=\frac{32-17}{5}=3$

d=10−5T10​−T5​​=532−17​=3.

Question: If the sum of 4th and 9th terms of an AP is 60 and the common difference is 3, find the first term.

Answer: Using

${𝑇}_4=𝑎+3(4-1)$

T4​=a+3(4−1) and

${𝑇}_9=𝑎+3(9-1)$

T9​=a+3(9−1), we get

$𝑎=8$

a=8.

Question: If the 12th term of an AP is 29 and the common difference is 2, find the 7th term.

Answer: Using

${𝑇}_7=𝑎+6𝑑$

T7​=a+6d, we get

${𝑇}_7=29-10=19$

T7​=29−10=19.

Question: If the first term of an AP is 4 and the 8th term is 22, find the common difference.

Answer:

$𝑑=\frac{{𝑇}_8-𝑎}{8-1}=\frac{22-4}{7}=2$

d=8−1T8​−a​=722−4​=2.

Question: What is the formula for the nth term of an AP in terms of the first term, common difference, and number of terms?

Answer:

${𝑇}_{𝑛}=𝑎+(𝑛-1)𝑑$

Tn​=a+(n−1)d.

Question: How can you determine if a given sequence of numbers is an AP?

Answer: Check if the difference between consecutive terms is constant. If it is, the sequence is an AP.

Question: What is the sum of an AP if its first term is 3, common difference is 4, and it has 10 terms?

Answer:

${𝑆}_{10}=\frac{10}{2}(2(3)+(10-1)(4))=245$

S10​=210​(2(3)+(10−1)(4))=245.

Also Check: Surface Areas and Volumes Class 10 Extra Questions Maths Chapter 13

Question: If the sum of first ‘n’ terms of an AP is

${𝑆}_{𝑛}$

Sn​, what is

${𝑆}_{𝑛}-{𝑆}_{𝑛-1}$

Sn​−Sn−1​?

Answer:

${𝑆}_{𝑛}-{𝑆}_{𝑛-1}={𝑇}_{𝑛}$

Sn​−Sn−1​=Tn​.

Question: What is the sum of all odd terms in an AP?

Answer:

${𝑆}_{\text{odd}}=\frac{𝑛}{2}(2𝑎+(𝑛-1)𝑑)$

Sodd​=2n​(2a+(n−1)d) where ‘n’ is the number of odd terms.

Question: How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78?

Answer: Given AP: 24, 21, 18,…

Here, a = 24, d = 21 – 24 = –3, S_n = 78. We need to find n.
We know that;

S_n = n/2[2a + (n – 1)d]

So, 78 = n/2 [48 + (n – 1)(-3)]

78 = n/2 [51 – 3n]

156 = 51n – 3n²

3n² – 51n + 156 = 0

n² – 17n + 52 = 0

n² – 13n – 4n + 52 = 0

n(n – 13) – 4(n – 13) = 0

(n – 4) (n – 13) = 0

n = 4 or 13
Both values of n are admissible. So, the number of terms is either 4 or 13.

Question: If the 3rd term of an AP is 8 and the 6th term is 20, find the 10th term.

Answer: Using

${𝑇}_3=𝑎+2𝑑$

T3​=a+2d and

${𝑇}_6=𝑎+5𝑑$

T6​=a+5d, we find

${𝑇}_{10}=34$

T10​=34.

Question: What is the 30th term of an AP if the 10th term is 20 and the common difference is 3?

Answer:

${𝑇}_{30}=20+20(3)=80$

T30​=20+20(3)=80.

Question: How can you find the common difference if given the first and third terms of an AP?

Answer:

$𝑑={𝑇}_3-𝑎$

d=T3​−a.

Question: If the sum of the first 5 terms of an AP is 60 and the sum of the first 10 terms is 210, find the sum of the next 5 terms.

Answer:

${𝑆}_{10}-{𝑆}_5=210-60=150$

S10​−S5​=210−60=150.

Question: What is the 12th term of an AP if the 15th term is 55 and the common difference is 3?

Answer:

${𝑇}_{12}=55-3(15-12)=46$

T12​=55−3(15−12)=46.

Question: If the 4th term of an AP is 9 and the 7th term is 15, find the first term and common difference.

Answer: Using

${𝑇}_4=𝑎+3𝑑$

T4​=a+3d and

${𝑇}_7=𝑎+6𝑑$

T7​=a+6d, we find

$𝑎=3$

a=3 and

$𝑑=2$

d=2.