• NCERT Solutions
    • NCERT Library
  • RD Sharma
    • RD Sharma Class 12 Solutions
    • RD Sharma Class 11 Solutions Free PDF Download
    • RD Sharma Class 10 Solutions
    • RD Sharma Class 9 Solutions
    • RD Sharma Class 8 Solutions
    • RD Sharma Class 7 Solutions
    • RD Sharma Class 6 Solutions
  • Class 12
    • Class 12 Science
      • NCERT Solutions for Class 12 Maths
      • NCERT Solutions for Class 12 Physics
      • NCERT Solutions for Class 12 Chemistry
      • NCERT Solutions for Class 12 Biology
      • NCERT Solutions for Class 12 Economics
      • NCERT Solutions for Class 12 Computer Science (Python)
      • NCERT Solutions for Class 12 Computer Science (C++)
      • NCERT Solutions for Class 12 English
      • NCERT Solutions for Class 12 Hindi
    • Class 12 Commerce
      • NCERT Solutions for Class 12 Maths
      • NCERT Solutions for Class 12 Business Studies
      • NCERT Solutions for Class 12 Accountancy
      • NCERT Solutions for Class 12 Micro Economics
      • NCERT Solutions for Class 12 Macro Economics
      • NCERT Solutions for Class 12 Entrepreneurship
    • Class 12 Humanities
      • NCERT Solutions for Class 12 History
      • NCERT Solutions for Class 12 Political Science
      • NCERT Solutions for Class 12 Economics
      • NCERT Solutions for Class 12 Sociology
      • NCERT Solutions for Class 12 Psychology
  • Class 11
    • Class 11 Science
      • NCERT Solutions for Class 11 Maths
      • NCERT Solutions for Class 11 Physics
      • NCERT Solutions for Class 11 Chemistry
      • NCERT Solutions for Class 11 Biology
      • NCERT Solutions for Class 11 Economics
      • NCERT Solutions for Class 11 Computer Science (Python)
      • NCERT Solutions for Class 11 English
      • NCERT Solutions for Class 11 Hindi
    • Class 11 Commerce
      • NCERT Solutions for Class 11 Maths
      • NCERT Solutions for Class 11 Business Studies
      • NCERT Solutions for Class 11 Accountancy
      • NCERT Solutions for Class 11 Economics
      • NCERT Solutions for Class 11 Entrepreneurship
    • Class 11 Humanities
      • NCERT Solutions for Class 11 Psychology
      • NCERT Solutions for Class 11 Political Science
      • NCERT Solutions for Class 11 Economics
      • NCERT Solutions for Class 11 Indian Economic Development
  • Class 10
    • NCERT Solutions for Class 10 Maths
    • NCERT Solutions for Class 10 Science
    • NCERT Solutions for Class 10 Social Science
    • NCERT Solutions for Class 10 English
    • NCERT Solutions For Class 10 Hindi Sanchayan
    • NCERT Solutions For Class 10 Hindi Sparsh
    • NCERT Solutions For Class 10 Hindi Kshitiz
    • NCERT Solutions For Class 10 Hindi Kritika
    • NCERT Solutions for Class 10 Sanskrit
    • NCERT Solutions for Class 10 Foundation of Information Technology
  • Class 9
    • NCERT Solutions for Class 9 Maths
    • NCERT Solutions for Class 9 Science
    • NCERT Solutions for Class 9 Social Science
    • NCERT Solutions for Class 9 English
    • NCERT Solutions for Class 9 Hindi
    • NCERT Solutions for Class 9 Sanskrit
    • NCERT Solutions for Class 9 Foundation of IT
  • CBSE Sample Papers
    • Previous Year Question Papers
    • CBSE Topper Answer Sheet
    • CBSE Sample Papers for Class 12
    • CBSE Sample Papers for Class 11
    • CBSE Sample Papers for Class 10
    • Solved CBSE Sample Papers for Class 9 with Solutions 2024-2025
    • CBSE Sample Papers Class 8
    • CBSE Sample Papers Class 7
    • CBSE Sample Papers Class 6
  • Textbook Solutions
    • Lakhmir Singh
    • Lakhmir Singh Class 10 Physics
    • Lakhmir Singh Class 10 Chemistry
    • Lakhmir Singh Class 10 Biology
    • Lakhmir Singh Class 9 Physics
    • Lakhmir Singh Class 9 Chemistry
    • PS Verma and VK Agarwal Biology Class 9 Solutions
    • Lakhmir Singh Science Class 8 Solutions

Learn CBSE

NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction

April 24, 2019 by Sastry CBSE

NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction.

NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction

Short Answer Type Questions
Q1. Give an example of a statement P(n) which is true for all n≥ 4 but P(l), P(2) and P(3) are not true. Justify your answer.

Sol. Consider the statement P(n): 3n < n!

For n = 1, 3 x 1 < 1!, which is not true
For n = 2, 3 x 2 < 2!, which is not true
For n = 3, 3 x 3 < 3!, which is not true
For n = 4, 3 x 4 < 4!, which is true
For n = 5, 3 x 5 < 5!, which is true

Q2. Give an example of a statement P(n) which is true for all Justify your answer.
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 1

Instruction for Exercises 3-16: Prove each of the statements in these Exercises by the Principle of Mathematical Induction.

Q3. 4n – 1 is divisible by 3, for each natural number
Sol: Let P(n): 4n – 1 is divisible by 3 for each natural number n.
Now, P(l): 41 – 1 = 3, which is divisible by 3 Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 4k – 1 is divisible by 3
or               4k – 1 = 3m, m∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 4k+1 – 1
= 4k-4-l
= 4(3m + 1) – 1  [Using (i)]
= 12 m + 3
= 3(4m + 1), which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q4. 23n – 1 is divisible by 7, for all natural numbers
Sol: Let P(n): 23n – 1 is divisible by 7
Now, P( 1): 23 — 1 = 7, which is divisible by 7.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 23k – 1 is divisible by 7.
or               23k -1 = 7m, m∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 23(k+1)– 1
= 23k.23– 1
= 8(7 m + 1) – 1
= 56 m + 7
= 7(8m + 1), which is divisible by 7.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q5. n3 – 7n + 3 is divisible by 3, for all natural numbers
Sol: Let P(n): n3 – 7n + 3 is divisible by 3, for all natural numbers n.
Now P(l): (l)3 – 7(1) + 3 = -3, which is divisible by 3.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k) = K3 – 7k + 3 is divisible by 3
or K3 – 7k + 3 = 3m, m∈ N         (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1 ):(k + l)3 – 7(k + 1) + 3
= k3 + 1 + 3k(k + 1) – 7k— 7 + 3 = k3 -7k + 3 + 3k(k + l)-6
= 3m + 3[k(k+l)-2]  [Using (i)]
= 3[m + (k(k + 1) – 2)], which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q6. 32n – 1 is divisible by 8, for all natural numbers
Sol: Let P(n): 32n – 1 is divisible by 8, for all natural numbers n.
Now, P(l): 32 – 1 = 8, which is divisible by 8.
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): 32k – 1 is divisible by 8
or               32k -1 = 8m, m ∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 32(k+1)– l
= 32k • 32 — 1
= 9(8m + 1) – 1     (using (i))
= 72m + 9 – 1
= 72m + 8
= 8(9m +1), which is divisible by 8 Thus P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q7. For any natural number n, 7n – 2n is divisible by 5.
Sol: Let P(n): 7n – 2n is divisible by 5, for any natural number n.
Now, P(l) = 71-21 = 5, which is divisible by 5.
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.

.’.  P(k) = 7k -2k is divisible by 5
or  7k – 2k = 5m, m∈ N                                                                           (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 7k+1 -2k+1
= 7k-7-2k-2
= (5 + 2)7k -2k-2
= 5.7k + 2.7k-2-2k
= 5.7k + 2(7k – 2k)
= 5 • 7k + 2(5 m)     (using (i))
= 5(7k + 2m), which divisible by 5.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q8. For any natural number n, xn -yn is divisible by x -y, where x and y are any integers with x ≠y
Sol:
Let P(n) : xn – yn is divisible by x – y, where x and y are any integers with x≠y.
Now, P(l): x1 -y1 = x-y, which is divisible by (x-y)
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): xk -yk is divisible by (x – y)
or   xk-yk = m(x-y),m ∈ N …(i)
Now, we have to prove that P(k + 1) is true.
P(k+l):xk+l-yk+l
= xk-x-xk-y + xk-y-yky
= xk(x-y) +y(xk-yk)
= xk(x – y) + ym(x – y)  (using (i))
= (x -y) [xk+ym], which is divisible by (x-y)
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q9. n3 -n is divisible by 6, for each natural number n≥
Sol: Let P(n): n3 – n is divisible by 6, for each natural number n> 2.
Now, P(2): (2)3 -2 = 6, which is divisible by 6.
Hence, P(2) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): k3 – k is divisible by 6
or    k3 -k= 6m, m∈ N       (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): (k+ l)3-(k+ 1)
= k3+ 1 +3k(k+ l)-(k+ 1)
= k3+ 1 +3k2 + 3k-k- 1 = (k3-k) + 3k(k+ 1)
= 6m + 3 k(k +1)  (using (i))
Above is divisible by 6.   (∴ k(k + 1) is even)
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n,n≥ 2.

Q10. n(n2 + 5) is divisible by 6, for each natural number
Sol: Let P(n): n(n2 + 5) is divisible by 6, for each natural number.
Now P(l): 1 (l2 + 5) = 6, which is divisible by 6.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): k( k2 + 5) is divisible by 6.
or K (k2+ 5) = 6m, m∈ N         (i)
Now, we have to prove that P(k + 1) is true.
P(K+l):(K+l)[(K+l)2 + 5]
= (K + l)[K2 + 2K+6]
= K3 + 3 K2 + 8K + 6
= (K2 + 5K) + 3 K2 + 3K + 6 =K(K2 + 5) + 3(K2 + K + 2)
= (6m) + 3(K2 + K + 2)        (using (i))
Now, K2 + K + 2 is always even if A is odd or even.
So, 3(K2 + K + 2) is divisible by 6 and hence, (6m) + 3(K2 + K + 2) is divisible by 6.
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q11. n2 < 2n, for all natural numbers n ≥
Sol: Let P(n): n2 < 2n for all natural numbers n≥ 5.
Now P(5): 52 < 25 or 25 < 32, which is true.
Hence, P(5) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k): k2 < 2k  (i)
Now, to prove that P(k + 1) is true, we have to show that P(k+ 1): (k+ l)2 <2k+1
Using (i), we get
(k + l)2 = k2 + 2k + 1 < 2k + 2k + 1         (ii)
Now let, 2k + 2k + 1 < 2k+1     (iii)
∴ 2k + 2k + 1 < 2 • 2k
2k + 1 < 2k, which is true for all k > 5 Using (ii) and (iii), we get (k + l)2 < 2k+1 Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n,n≥ 5.

Q12. 2n<(n + 2)! for all natural numbers
Sol: Let P(n): 2n < (n + 2)! for all natural numbers n.
P( 1): 2 < (1 + 2)! or 2 < 3! or 2 < 6, which is true.
Hence,P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k) :2k<(k + 2)!  (i)
To prove that P(k + 1) is true, we have to show that
P(k + 1): 2(k+ 1) < (k + 1 + 2)!
or 2(k+ 1) < (k + 3)!
Using (i), we get
2(k + 1) = 2k + 2<(k+2) !  +2  (ii)
Now let, (k + 2)! + 2 < (k + 3)!  (iii)
=>  2 < (k+ 3)! – (k+2) !
=> 2 < (k + 2) ! [k+ 3-1]
=>2<(k+ 2) ! (k + 2), which is true for any natural number.
Using (ii) and (iii), we get 2(k + 1) < (k + 3)!
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 2
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 3

Q14. 2 + 4 + 6+… + 2n = n2 + n, for all natural numbers
Sol: Let P(n) :2 + 4 + 6+ …+2 n = n2 + n
P(l): 2 = l2 + 1 = 2, which is true
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k): 2 + 4 + 6 + .,.+2k = k2 + k  (i)
Now, we have to prove that P(k + 1) is true.
P(k + l):2 + 4 + 6 + 8+ …+2k+ 2 (k +1)
= k2 + k + 2(k+ 1)  [Using (i)]
= k2 + k + 2k + 2
= k2 + 2k+1+k+1
= (k + 1)2 + k+ 1
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q15. 1 + 2 + 22 + … + 2n = 2n +1 – 1 for all natural numbers
Sol: Let P(n): 1 + 2 + 22 + … + 2n = 2n +1 – 1, for all natural numbers n
P(1): 1 =20 + 1 — 1 = 2 — 1 = 1, which is true.
Hence, ,P(1) is true.
Let us assume that P(n) is true for some natural number n = k.

P(k): l+2 + 22+…+2k = 2k+1-l              (i)

Now, we have to prove that P(k + 1) is true.

P(k+1): 1+2 + 22+ …+2k + 2k+1
= 2k +1 – 1 + 2k+1  [Using (i)]
= 2.2k+l– 1 = 1
= 2(k+1)+1-1
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q16. 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers
Sol: Let P(n): 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers n.
P(1): 1 = 1(2 x 1 – 1) = 1, which is true.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k):l+5 + 9 +…+(4k-3) = k(2k-1)  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 1 + 5 + 9 + … +  (4k- 3) + [4(k+ 1) – 3]
= 2k2 -k+4k+ 4-3
= 2k2 + 3k + 1
= (k+ 1)( 2k + 1)

= (k+l)[2(k+l)-l]

Hence, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for any natural number n.

Long Answer Type Questions
Q17. A sequence ax, a2, a3, … is defined by letting a1=3 and ak = 7ak–1 for all natural numbers k≥ Show that an = 3 • 7 n-1 for all natural numbers.
Sol: We have a sequence ax, a2, a3… defined by letting a, = 3 and ak = 7ak–1, for all natural numbers k≥2.
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 4

NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 5

Q18. A sequence b0, b1, b2, … is defined by letting b0 = 5 and bk = 4 + bk–1, for all natural numbers Show that bn = 5 + 4n, for all natural number n using mathematical induction.
Sol. We have a sequence b0, b1, b2,… defined by letting b0 = 5 and bk = 4 + bk–1,, for all natural numbers k.

NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 6
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 7
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 8
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 9
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 10
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 11
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 12
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 13
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 14
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 15
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 16
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 17
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 18
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 19
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 20
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 21
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 22
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 23
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 24
NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction Img 25

So, by the principle of mathematical induction P(n) is true for any natural number rt,n> 1.

Q25. Prove that number of subsets of a set containing n distinct elements is 2″, for all n ∈
Sol: Let P(n): Number of subset of a set containing n distinct elements is 2″, for all ne N.
For n = 1, consider set A = {1}. So, set of subsets is {{1}, ∅}, which contains 21 elements.
So, P(1) is true.
Let us assume that P(n) is true, for some natural number n = k.
P(k): Number of subsets of a set containing k distinct elements is 2k To prove that P(k + 1) is true,
we have to show that P(k + 1): Number of subsets of a set containing (k + 1) distinct elements is 2k+1
We know that, with the addition of one element in the set, the number of subsets become double.
Number of subsets of a set containing (k+ 1) distinct elements = 2×2k = 2k+1
So, P(k + 1) is true. Hence, P(n) is true.

NCERT Exemplar Class 11 Maths Solutions

  • Chapter 1 Sets
  • Chapter 2 Relations and Functions
  • Chapter 3 Trigonometric Functions
  • Chapter 4 Principle of Mathematical Induction
  • Chapter 5 Complex Numbers and Quadratic Equations
  • Chapter 6 Linear Inequalities
  • Chapter 7 Permutations and Combinations
  • Chapter 8 Binomial Theorem
  • Chapter 9 Sequence and Series
  • Chapter 10 Straight Lines
  • Chapter 11 Conic Sections
  • Chapter 12 Introduction to Three-Dimensional Geometry
  • Chapter 13 Limits and Derivatives
  • Chapter 14 Mathematical Reasoning
  • Chapter 15 Statistics
  • Chapter 16 Probability

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

We hope the NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction, drop a comment below and we will get back to you at the earliest.

Filed Under: Class 11 Mathematics

LearnCBSE.in Student Education Loan
  • Student Nutrition - How Does This Effect Studies
  • Words by Length
  • NEET MCQ
  • Factoring Calculator
  • Rational Numbers
  • CGPA Calculator
  • TOP Universities in India
  • TOP Engineering Colleges in India
  • TOP Pharmacy Colleges in India
  • Coding for Kids
  • Math Riddles for Kids with Answers
  • General Knowledge for Kids
  • General Knowledge
  • Scholarships for Students
  • NSP - National Scholarip Portal
  • Class 12 Maths NCERT Solutions
  • Class 11 Maths NCERT Solutions
  • NCERT Solutions for Class 10 Maths
  • NCERT Solutions for Class 9 Maths
  • NCERT Solutions for Class 8 Maths
  • NCERT Solutions for Class 7 Maths
  • NCERT Solutions for Class 6 Maths
  • NCERT Solutions for Class 6 Science
  • NCERT Solutions for Class 7 Science
  • NCERT Solutions for Class 8 Science
  • NCERT Solutions for Class 9 Science
  • NCERT Solutions for Class 10 Science
  • NCERT Solutions for Class 11 Physics
  • NCERT Solutions for Class 11 Chemistry
  • NCERT Solutions for Class 12 Physics
  • NCERT Solutions for Class 12 Chemistry
  • NCERT Solutions for Class 10 Science Chapter 1
  • NCERT Solutions for Class 10 Science Chapter 2
  • Metals and Nonmetals Class 10
  • carbon and its compounds class 10
  • Periodic Classification of Elements Class 10
  • Life Process Class 10
  • NCERT Solutions for Class 10 Science Chapter 7
  • NCERT Solutions for Class 10 Science Chapter 8
  • NCERT Solutions for Class 10 Science Chapter 9
  • NCERT Solutions for Class 10 Science Chapter 10
  • NCERT Solutions for Class 10 Science Chapter 11
  • NCERT Solutions for Class 10 Science Chapter 12
  • NCERT Solutions for Class 10 Science Chapter 13
  • NCERT Solutions for Class 10 Science Chapter 14
  • NCERT Solutions for Class 10 Science Chapter 15
  • NCERT Solutions for Class 10 Science Chapter 16

Free Resources

RD Sharma Class 12 Solutions RD Sharma Class 11
RD Sharma Class 10 RD Sharma Class 9
RD Sharma Class 8 RD Sharma Class 7
CBSE Previous Year Question Papers Class 12 CBSE Previous Year Question Papers Class 10
NCERT Books Maths Formulas
CBSE Sample Papers Vedic Maths
NCERT Library

NCERT Solutions

NCERT Solutions for Class 10
NCERT Solutions for Class 9
NCERT Solutions for Class 8
NCERT Solutions for Class 7
NCERT Solutions for Class 6
NCERT Solutions for Class 5
NCERT Solutions for Class 4
NCERT Solutions for Class 3
NCERT Solutions for Class 2
NCERT Solutions for Class 1

Quick Resources

English Grammar Hindi Grammar
Textbook Solutions Maths NCERT Solutions
Science NCERT Solutions Social Science NCERT Solutions
English Solutions Hindi NCERT Solutions
NCERT Exemplar Problems Engineering Entrance Exams
Like us on Facebook Follow us on Twitter
Watch Youtube Videos NCERT Solutions App