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,
an = 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.
an = 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
of an AP is given by the formula
.
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,
an = 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
, where
is the first term,
is the last term, and
is the common difference.
Question: What is the sum of the first ‘n’ terms of an AP?
Answer: The sum
of the first ‘n’ terms of an AP is given by the formula
.
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:
.
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
and
, we get
.
Question: If the 12th term of an AP is 29 and the common difference is 2, find the 7th term.
Answer: Using
, we get
.
Question: If the first term of an AP is 4 and the 8th term is 22, find the common difference.
Answer:
.
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:
.
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:
.
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
, what is
?
Answer:
.
Question: What is the sum of all odd terms in an AP?
Answer:
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, Sn = 78. We need to find n.
We know that;
Sn = 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
and
, we find
.
Question: What is the 30th term of an AP if the 10th term is 20 and the common difference is 3?
Answer:
.
Question: How can you find the common difference if given the first and third terms of an AP?
Answer:
.
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:
.
Question: What is the 12th term of an AP if the 15th term is 55 and the common difference is 3?
Answer:
.
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
and
, we find
and
.