If f is a function satisfying | 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 5, Question 7:

If f is a function satisfying f(x +y) = f(x) f(y) for all x,y element ofN  such that f(1) = 3

sum from x equals 1 to n of space f left parenthesis x right parenthesis space equals space 120 and  , find the value of n.

. 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 7:

If f is a function satisfying f(x +y) = f(x) f(y) for all x,y element ofN  such that f(1) = 3

sum from x equals 1 to n of space f left parenthesis x right parenthesis space equals space 120 and  , find the value of n.

Answer:

It is given that,

f (x + y) = f (x) × f (y) for all x, y ∈ N … (1)

f (1) = 3

Taking x = y = 1 in (1), we obtain

f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9

Similarly,

f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27

f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81

∴ f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with both the first term and common ratio equal to 3.

It is known that, S subscript n space equals space fraction numerator a open parentheses r to the power of n space minus space 1 close parentheses over denominator r space minus 1 end fraction

It is given that, sum from x equals 1 to n of space f left parenthesis x right parenthesis space equals space 120

therefore space 120 space equals space fraction numerator 3 open parentheses 3 to the power of n space minus 1 close parentheses over denominator 3 minus 1 end fraction
rightwards double arrow 120 space equals space 3 over 2 open parentheses 3 to the power of n space minus space 1 close parentheses
rightwards double arrow 3 to the power of n space minus space 1 space equals space 80
rightwards double arrow 3 to the power of n space equals space 81 space equals space 3 to the power of 4 space
therefore space n space equals 4

Thus, the value of n is 4.


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 5 , Question 7: If f is a function satisfying f(x +y) = f(x) f(y) for all x,y N  such that f(1) ....

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