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Example 14 - Find sum of (i) first 1000 positive integers - Finding sum of n terms

  1. Chapter 5 Class 10 Arithmetic Progressions
  2. Serial order wise
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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, โ€ฆโ€ฆโ€ฆ. 1000 We have to find the sum , So, We use the formula Sum = ๐‘›/2[2๐‘Ž+(๐‘›โˆ’1)๐‘‘] Here, a = 1 d = 2 โ€“ 1 = 1 n = 1000 Putting these values in formula Sum = ๐‘›/2[2๐‘Ž+(๐‘›โˆ’1)๐‘‘] Sum = 1000/2[2ร—1+(1000โˆ’1)(1)] Sum = 1000/2[2+999] Sum = 500 ร—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, โ€ฆโ€ฆโ€ฆ.1000 We have to find the sum , We use the formula Sum = ๐‘›/2[๐‘Ž+๐‘™] Given last term = ๐‘™ = 1000 And a = 1 , n = 1000 Putting these values in formula Sum = ๐‘›/2[๐‘Ž+๐‘™] Sum = 1000/2[1+1000] Sum = 500 ร—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 โ€ฆโ€ฆโ€ฆโ€ฆ.n We have to find the sum of the series, So, we use the formula Sum = ๐‘›/2(๐‘Ž+๐‘™) where a = 1 & last term = ๐‘™ = n & n = n Putting these values in formula Sum = ๐‘›/2(๐‘Ž+๐‘™) Sum = ๐‘›/2 (1+๐‘›) Sum = (๐‘›(๐‘› + 1))/2

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