Question 45 of 51intermediate🔧 ApplyNumerical4 marks

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

Correct Answer

Step 1: The formula for the sum of the first n natural numbers is S = n(n + 1) / 2.

Step 2: We need the sum to be greater than 1,000, so n(n + 1) / 2 > 1000.

Step 3: Multiply both sides by 2 to get n(n + 1) > 2000.

Step 4: Find the smallest natural number n satisfying this. For n = 44, 44×4544 \times 45 = 1980 (which is < 2000). For n = 45, 45×4645 \times 46 = 2070 (which is > 2000).

Step 5: Therefore, the smallest value of n is 45.

Exercise: END-OF-CHAPTER EXERCISES | Q: 9 | (Page 22)
For More Understanding

Explanation

The textbook provides the formula for the sum of the first n natural numbers as S = n(n + 1) / 2. To find the smallest n such that the sum is greater than 1000, we set up the inequality n(n + 1) / 2 > 1000, which simplifies to n(n + 1) > 2000. By checking consecutive natural numbers, we find that for n = 44, 44 * 45 = 1980 (not > 2000), but for n = 45, 45 * 46 = 2070 (> 2000). Thus, the smallest value of n is 45.

Solution Steps

  1. Step 1: The formula for the sum of the first n natural numbers is S = n(n + 1) / 2.

  2. Step 2: We need S > 1000, so n(n + 1) / 2 > 1000.

  3. Step 3: Multiply both sides by 2 to get n(n + 1) > 2000.

  4. Step 4: Test values of n. For n = 44, 44×4544 \times 45 = 1980 (< 2000). For n = 45, 45×4645 \times 46 = 2070 (> 2000).

  5. Step 5: Therefore, the smallest value of n is 45.

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