Learn all Concepts of Chapter 5 Class 10 (with VIDEOS). Check - Arithmetic Progressions - Class 10

Last updated at March 19, 2021 by Teachoo

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 Example 14 Find the sum of : (ii) the first n positive integers Positive integers start from 1 First n positive integers are 1, 2, 3, 4, ………., n This is an AP with First term = a = 1 Common Difference = d = 1 Number of terms = n = (n − 1) + 1 = n Last term = l = n To find sum, we use the formula Sum = 𝒏/𝟐[𝒂+𝒍] Putting values Sum = 𝑛/2 (1+𝑛) Sum = (𝑛(𝒏 + 𝟏))/2

Chapter 5 Class 10 Arithmetic Progressions

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.