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NCERT Solutions for Class 11 Mathematics Chapter 7 Permutations And Combinations
  • NCERT Solution For Class 11 Mathematics

    Permutations And Combinations Here is the CBSE Mathematics Chapter 7 for Class 11 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 11 Mathematics Permutations And Combinations Chapter 7 NCERT Solutions for Class 11 Mathematics Permutations And Combinations Chapter 7 The following is a summary in Hindi and English for the academic year 2024-25. You can save these solutions to your computer or use the Class 11 Mathematics.

    Question 1
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    Write the first five terms of each of the sequences whose nth terms are :

    straight a subscript straight n space equals space fraction numerator straight n left parenthesis straight n squared plus 5 right parenthesis over denominator 4 end fraction

    Solution

    Here straight a subscript straight n space equals space fraction numerator straight n left parenthesis straight n squared plus 5 right parenthesis over denominator 4 end fraction
    Putting n = 1, 2, 3, 4, 5, we get
            space space straight a subscript 1 space equals space fraction numerator 1 left parenthesis 1 squared plus 5 right parenthesis over denominator 4 end fraction space equals space 6 over 4 space equals space 3 over 2 comma space space space straight a subscript 2 space equals space fraction numerator 2 left parenthesis 2 squared plus 5 right parenthesis over denominator 4 end fraction space equals space 9 over 2 comma space space straight a subscript 3 space equals space fraction numerator 3 left parenthesis 3 squared plus 5 right parenthesis over denominator 4 end fraction space equals space 21 over 2
              straight a subscript 4 equals space fraction numerator 4 left parenthesis 4 squared plus 5 right parenthesis over denominator 4 end fraction equals 21 comma space space space straight a subscript 5 equals space fraction numerator 5 left parenthesis 5 squared plus 5 right parenthesis over denominator 4 end fraction space equals space 75 over 2

    ∴ First five terms are 3 over 2 comma space space 9 over 2 comma space 21 over 2 comma space 21 comma space 75 over 2

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