The sum of k consecutive natural numbers is k(2a + k – 1) / 2 = 100, meaning k(2a + k – 1) = 200. Testing odd factors gives only k = 5, yielding 18 + 19 + 20 + 21 + 22 = 100.
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The sum of k consecutive natural numbers is k(2a + k – 1) / 2 = 100, meaning k(2a + k – 1) = 200. Testing odd factors gives only k = 5, yielding 18 + 19 + 20 + 21 + 22 = 100.
Let k consecutive natural numbers start at a, where k >= 2.
Their sum is:
S = (k / 2) x [2a + (k – 1)] = 100
k(2a + k – 1) = 200.
Here, one factor must be odd and the other even. The odd factors of 200 are 1, 5 and 25.
For k = 5:
5(2a + 4) = 200
2a + 4 = 40
2a = 36, so a = 18.
For k = 25:
25(2a + 24) = 200 gives 2a + 24 = 8, so a = -8 (not natural).
Thus, the only way is 18 + 19 + 20 + 21 + 22.
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/