Which term of the following sequences: | 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 3, Question 5:

Which term of the following sequences:

2 comma space 2 square root of 2 comma 4 comma space...

. 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 5:

Which term of the following sequences:

2 comma space 2 square root of 2 comma 4 comma space...

Answer:

(a) The given sequence is 2 comma space 2 square root of 2 comma 4 comma space...

Here, a = 2 and r = fraction numerator 2 square root of 2 over denominator 2 end fraction equals square root of 2

Let the nth term of the given sequence be 128.

a subscript n space end subscript equals space a r to the power of n minus 1 end exponent
rightwards double arrow open parentheses 2 close parentheses open parentheses square root of 2 close parentheses to the power of n minus 1 end exponent space equals space 128

rightwards double arrow open parentheses 2 close parentheses open parentheses 2 close parentheses to the power of fraction numerator n minus 1 over denominator 2 end fraction end exponent equals open parentheses 2 close parentheses to the power of 7

rightwards double arrow open parentheses 2 close parentheses to the power of fraction numerator n minus 1 over denominator 2 end fraction plus 1 end exponent equals open parentheses 2 close parentheses to the power of 7

s o space fraction numerator n minus 1 over denominator 2 end fraction plus 1 equals 7
rightwards double arrow fraction numerator n minus 1 over denominator 2 end fraction space equals space 6
rightwards double arrow n space minus 1 space equals space 12
rightwards double arrow n equals 13

Thus, the 13th term of the given sequence is 128.

(b) The given sequence is square root of 3 space comma space 3 comma space 3 square root of 3 space comma space... space i s space 729 ?

Here, a= square root of 3 space a n d space r space equals space fraction numerator 3 over denominator square root of 3 end fraction space equals space square root of 3

Let the nth term of the given sequence be 729.

a subscript n equals a r to the power of n minus 1 end exponent
s o
a r to the power of n minus 1 end exponent space equals space 729
rightwards double arrow open parentheses square root of 3 close parentheses space open parentheses square root of 3 close parentheses to the power of n minus 1 end exponent space equals space 729

rightwards double arrow open parentheses 3 close parentheses to the power of 1 half end exponent open parentheses 3 close parentheses to the power of fraction numerator n minus 1 over denominator 2 end fraction end exponent equals open parentheses 3 close parentheses to the power of 6

rightwards double arrow open parentheses 3 close parentheses to the power of 1 half plus fraction numerator n minus 1 over denominator 2 end fraction end exponent equals open parentheses 3 close parentheses to the power of 6
s o
1 half plus fraction numerator n minus 1 over denominator 2 end fraction equals 6
rightwards double arrow fraction numerator 1 plus n minus 1 over denominator 2 end fraction equals 6
rightwards double arrow n equals 12

Thus, the 12th term of the given sequence is 729.

(c) The given sequence is 1 third comma space 1 over 9 comma space 1 over 27 comma space... space

Here, 

a=1 third space space a n d space r space equals space 1 over 9 obelus divided by 1 third equals 1 third

Let the nth term of the given sequence be 1 over 19683.

a subscript n space equals space a r to the power of n minus 1 end exponent
s o
a r to the power of n minus 1 end exponent equals 1 over 19683
rightwards double arrow open parentheses 1 third close parentheses open parentheses 1 third close parentheses to the power of n minus 1 end exponent equals 1 over 19683
rightwards double arrow open parentheses 1 third close parentheses to the power of n equals open parentheses 1 third close parentheses to the power of 9
rightwards double arrow n equals 9

Thus, the 9th term of the given sequence is 1 over 19683.


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 3 , Question 5: Which term of the following sequences:....

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