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Srushti
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Asked: 2026-09-10T05:36:55+00:00 In:

Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.

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The sum formula is n(n + 1) / 2 > 1000, so n(n + 1) > 2000. Testing values gives 44 x 45 / 2 = 990 and 45 x 46 / 2 = 1035, so smallest n is 45.

000cbse class 9 ganita manjari chapter 8 exercise 8.3 solutionsclass 9 ganita manjari chapter 8 step by step solutionssmallest value of n such that the sum of the first n natural numberssum of the first n natural numbers is greater than 1
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    1. The sum of the first n natural numbers is given by:

      Sn = n(n + 1) / 2

      We need the smallest n such that:

      n(n + 1) / 2 > 1000

      n(n + 1) > 2000.

      Since 44 x 44 = 1936 and 45 x 45 = 2025,

      let us check consecutive integers near 44:

      For n = 44:

      S44 = 44 x 45 / 2 = 22 x 45 = 990 (less than 1000).

      For n = 45:

      S45 = 45 x 46 / 2 = 45 x 23 = 1035 (greater than 1000).

      Thus, the smallest value of n is 45.

       

      For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)

      https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/

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