Example 14 (i) - Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Last updated at Aug. 3, 2021 by Teachoo
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Example 14 (i) You are here
Example 14 (ii)
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Example 16 Important Deleted for CBSE Board 2022 Exams
Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Serial order wise
Transcript
Example 14 (Method 1) Find the sum of the first 1000 positive integers Positive integers start from 1 First 1000 positive integers are 1, 2, 3, 4, ………., 1000 This is an AP with First term = a = 1 Common Difference = d = 1 Number of terms = n = (1000 − 1) + 1 = 1000 To find sum, we use the formula Sum = 𝒏/𝟐[𝟐𝒂+(𝒏−𝟏)𝒅] Putting values Sum = 1000/2[2 × 1+(1000−1)(1)] Sum = 1000/2[2+999] Sum = 500 × 1001 Sum = 500500 Example 14 (Method 2) Find the sum of the first 1000 positive integers Positive integers start from 1 First 1000 positive integers are 1, 2, 3, 4, ………., 1000 This is an AP with First term = a = 1 Common Difference = d = 1 Number of terms = n = (1000 − 1) + 1 = 1000 Last term = l = 1000 To find sum, we use the formula Sum = 𝒏/𝟐[𝒂+𝒍] Putting values Sum = 1000/2[1+1000] Sum = 500 × 1001 Sum = 500500
