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

Last updated at Feb. 25, 2017 by Teachoo

Transcript

Example 14(Method 1) Find the sum of the first 1000 positive integers Sum of first 1000 positive integers . Positive integers start from 1. So, 1, 2, 3, 4, ………. 1000 We have to find the sum , So, We use the formula Sum = 𝑛/2[2𝑎+(𝑛−1)𝑑] Here, a = 1 d = 2 – 1 = 1 n = 1000 Putting these values in formula Sum = 𝑛/2[2𝑎+(𝑛−1)𝑑] 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 Sum of first 1000 positive integers . Positive integers start from 1. So, 1,2,3,4, ……….1000 We have to find the sum , We use the formula Sum = 𝑛/2[𝑎+𝑙] Given last term = 𝑙 = 1000 And a = 1 , n = 1000 Putting these values in formula Sum = 𝑛/2[𝑎+𝑙] Sum = 1000/2[1+1000] Sum = 500 ×1001 = 500500 Example 14 Find the sum of : (ii) the first n positive integers The sum of first n positive integers Here , positive integer would start from 1 and continue till n So, the series is 1, 2, 3, 4 ………….n We have to find the sum of the series, So, we use the formula Sum = 𝑛/2(𝑎+𝑙) where a = 1 & last term = 𝑙 = n & n = n Putting these values in formula Sum = 𝑛/2(𝑎+𝑙) 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 9 years. He provides courses for Maths and Science at Teachoo.