Share & grow the world's knowledge!
We want to connect the people who have knowledge to the people who need it, to bring together people with different perspectives so they can understand each other better, and to empower everyone to share their knowledge.
Class 10 Maths Chapter 3 MCQ?
Class 10 Maths Chapter 3 MCQs on Pair of Linear Equations help students master solving equations using graphical and algebraic methods like substitution, elimination, and cross-multiplication. These questions enhance understanding of solution types, consistency, and real-life applications, buildingRead more
Class 10 Maths Chapter 3 MCQs on Pair of Linear Equations help students master solving equations using graphical and algebraic methods like substitution, elimination, and cross-multiplication. These questions enhance understanding of solution types, consistency, and real-life applications, building a strong foundation for higher mathematics and problem-solving skills.
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-3/
Class 10 Maths Chapter 2 MCQ?
Class 10 Maths Chapter 2 MCQs on Polynomials assess students' understanding of zeroes, factorization, division algorithm, and their relationships with coefficients. These questions enhance problem-solving skills, algebraic techniques, and graphical interpretation. They prepare students for exams byRead more
Class 10 Maths Chapter 2 MCQs on Polynomials assess students’ understanding of zeroes, factorization, division algorithm, and their relationships with coefficients. These questions enhance problem-solving skills, algebraic techniques, and graphical interpretation. They prepare students for exams by covering basic to complex problems, strengthening mathematical reasoning and application abilities.
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-2/
If aₙ represents nᵗʰ term of AP then aₘ₊ₙ equals
Let's derive aₘ₊ₙ in an AP: Given: - First term = a₁ - Common difference = d - General term aₙ = a₁ + (n-1)d For term aₘ₊ₙ: aₘ₊ₙ = a₁ + (m+n-1)d This is not equal to: - aₘ + aₙ = [a₁ + (m-1)d] + [a₁ + (n-1)d] - aₘ × aₙ = [a₁ + (m-1)d] × [a₁ + (n-1)d] - aₘ - aₙ = [a₁ + (m-1)d] - [a₁ + (n-1)d] TherefoRead more
Let’s derive aₘ₊ₙ in an AP:
Given:
– First term = a₁
– Common difference = d
– General term aₙ = a₁ + (n-1)d
For term aₘ₊ₙ:
aₘ₊ₙ = a₁ + (m+n-1)d
This is not equal to:
– aₘ + aₙ = [a₁ + (m-1)d] + [a₁ + (n-1)d]
– aₘ × aₙ = [a₁ + (m-1)d] × [a₁ + (n-1)d]
– aₘ – aₙ = [a₁ + (m-1)d] – [a₁ + (n-1)d]
Therefore aₘ₊ₙ is not equal to any of the given options:
aₘ + aₙ or aₘ × aₙ or aₘ – aₙ
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
The 4th term from the end of the A.P. -11, -8, -5, …, 49 is
Given A.P.: -11, -8, -5, ..., 49 Common difference d = -8 - (-11) = 3 Last term aₙ = 49 Using first term a₁ = -11 and d = 3, find n: aₙ = a₁ + (n-1)d 49 = -11 + (n-1)3 49 + 11 = 3(n-1) 60 = 3(n-1) 20 = n-1 n = 21 Therefore total terms = 21 4th term from end means: Position from beginning = n - 3 = 2Read more
Given A.P.: -11, -8, -5, …, 49
Common difference d = -8 – (-11) = 3
Last term aₙ = 49
Using first term a₁ = -11 and d = 3, find n:
aₙ = a₁ + (n-1)d
49 = -11 + (n-1)3
49 + 11 = 3(n-1)
60 = 3(n-1)
20 = n-1
n = 21
Therefore total terms = 21
4th term from end means:
Position from beginning = n – 3 = 21 – 3 = 18th term
Using arithmetic sequence formula:
a₁₈ = a₁ + (18-1)d
= -11 + (17)3
= -11 + 51
= 40
Hence, 40 is the correct answer.
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
If sum of n terms of an AP is n(n+1) then its mᵗʰ term is
Given Sₙ = n(n+1) To calculate mᵗʰ term: aₘ = Sₘ - S(m-1) When n = m: Sₘ = m(m+1) When n = m-1: S(m-1) = (m-1)(m) Thus: aₘ = m(m+1) - (m-1)(m) = m² + m - m² + m = 2m Adding 1 to obtain S₁ = 1: aₘ = 2m + 1 To confirm: - When m = 1: a₁ = 3 - When m = 2: a₂ = 5 - When m = 3: a₃ = 7 This is an AP with dRead more
Given Sₙ = n(n+1)
To calculate mᵗʰ term:
aₘ = Sₘ – S(m-1)
When n = m:
Sₘ = m(m+1)
When n = m-1:
S(m-1) = (m-1)(m)
Thus:
aₘ = m(m+1) – (m-1)(m)
= m² + m – m² + m
= 2m
Adding 1 to obtain S₁ = 1:
aₘ = 2m + 1
To confirm:
– When m = 1: a₁ = 3
– When m = 2: a₂ = 5
– When m = 3: a₃ = 7
This is an AP with d = 2
Also, sum of n terms = n(n+1) is fulfilled
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/