Between 1 and 31, m numbers ha | Class 11 Mathematics Chapter Sequence and Series, Sequence and Series NCERT Solutions

Current NCERT chapter

Academic session 2026-27, Chapter 8 in the current curriculum.

Welcome to the NCERT Solutions for Class 11 Mathematics - Chapter Sequence and Series. This page offers a step-by-step solution to the specific question from Exercise 2, Question 16:

Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)thnumbers is 5:9. Find the value of m.

. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.

Question 16:

Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)thnumbers is 5:9. Find the value of m.

Answer:

Let A1, A2, … Am be m numbers such that 1, A1, A2, … Am, 31 is an A.P.

Here, a = 1, b = 31, n = m + 2

∴ 31 = 1 + (m + 2 – 1) (d)

⇒ 30 = (m + 1) d

rightwards double arrow d equals fraction numerator 30 over denominator m plus 1 end fraction space space space space space space space space... left parenthesis 1 right parenthesis

A1 = a + d

A2 = a + 2d

A3 = a + 3d …

∴ A7 = a + 7d

Am–1 = a + (m – 1) d

According to the given condition,

fraction numerator a plus 7 d over denominator a plus open parentheses m minus 1 close parentheses d end fraction equals 5 over 9

rightwards double arrow fraction numerator 1 plus 7 open parentheses begin display style fraction numerator 30 over denominator open parentheses m plus 1 close parentheses end fraction end style close parentheses over denominator 1 plus open parentheses m minus 1 close parentheses open parentheses begin display style fraction numerator 30 over denominator m plus 1 end fraction end style close parentheses end fraction equals 5 over 9 space space space space space space space space space space space space space space space space open square brackets F r o m space left parenthesis 1 right parenthesis close square brackets

rightwards double arrow fraction numerator m plus 1 plus 7 open parentheses 30 close parentheses over denominator m plus 1 plus 30 open parentheses m minus 1 close parentheses end fraction equals 5 over 9

rightwards double arrow fraction numerator m plus 1 plus 210 over denominator m plus 1 plus 30 m minus 30 end fraction equals 5 over 9

rightwards double arrow fraction numerator m plus 211 over denominator 31 m space minus 29 end fraction equals 5 over 9

rightwards double arrow 9 m space plus space 1899 space equals space 155 m space minus 145

rightwards double arrow 155 m space minus 9 m space equals space 1899 space plus 145

rightwards double arrow 146 m equals 2044
rightwards double arrow m equals 14

Thus, the value of m is 14.


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Sequence and Series.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

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Welcome to the NCERT Solutions for Class 11 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 16: Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence i....

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