Find the sum of integers from 1 to 100 t | Class 11 Mathematics Chapter Sequence and Series, Sequence and Series NCERT Solutions

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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 5:

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

. 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:

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Answer:

The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100.

This forms an A.P. with both the first term and common difference equal to 2.

⇒100 = 2 + (n –1) 2

⇒ n = 50

therefore space 2 plus 4 plus 6 plus... plus 100 space equals 50 over 2 space open square brackets 2 open parentheses 2 close parentheses space plus space open parentheses 50 minus 1 close parentheses open parentheses 2 close parentheses close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 50 over 2 space open square brackets 4 space plus space 98 close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses 25 close parentheses open parentheses 102 close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2550

The integers from 1 to 100, which are divisible by 5, are 5, 10… 100.

This forms an A.P. with both the first term and common difference equal to 5.

∴100 = 5 + (n –1) 5

⇒ 5n = 100

⇒ n = 20

therefore space 5 space plus space 10 space plus space... space plus space 100 space equals space 20 over 2 space open square brackets 2 open parentheses 5 close parentheses space plus space open parentheses 20 space minus 1 close parentheses 5 close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 10 space open square brackets 10 space plus space open parentheses 19 close parentheses 5 close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 10 space open square brackets 10 space plus space 95 close square brackets space equals space 10 space cross times space 105 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1050

The integers, which are divisible by both 2 and 5, are 10, 20, … 100.

This also forms an A.P. with both the first term and common difference equal to 10.

∴100 = 10 + (n –1) (10)

⇒ 100 = 10n

⇒ n = 10

therefore space 10 space plus space 20 space plus space... space plus space 100 space equals space 10 over 2 space open square brackets 2 open parentheses 10 close parentheses space plus space open parentheses 10 space minus 1 close parentheses open parentheses 10 close parentheses close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 5 space open square brackets 20 space plus space 90 close square brackets space equals space 5 open parentheses 110 close parentheses space equals space 550

∴Required sum = 2550 + 1050 – 550 = 3050

Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

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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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Comments

  • Nashitah Rahman
  • Jan 16, 2021

The answer was helpful but can we write directly the integers divisible by both 2 or 5??

  • Nashitah Rahman
  • Jan 16, 2021

The answer was helpful but can we write directly the integers divisible by both 2 or 5??

  • Priyanshu
  • Nov 13, 2019

Thanks for the solution.

  • Shivanshu
  • Oct 16, 2019

That is correct but I have asked other answer

  • Aarya
  • Sep 16, 2019

Great..

  • Deekshitha
  • Sep 16, 2019

Thanks sir it helped me

  • jitendra pushkar
  • Sep 13, 2019

thanks sir.. ... wonderful answer.....

  • Subhash
  • Sep 02, 2019

Thanks a lot

  • Ashutosh kumar
  • Aug 26, 2019

Thanks sir

  • Asha
  • Aug 23, 2019

Can it be solved by n(n+2)/2

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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 5: Find the sum of integers from 1 to 100 that are divisible by 2 or 5.....

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