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Question

# The series of natural numbers is divided into groups as follows ; (1), (2,3),(4,5,6), (7,8,9,10) and so on. The sum of the numbers in the Nth group is

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Solution

## In the mentioned query it can be seen that the number of natural numbers in the first, second, third,....., nthgroups are 1, 2, 3, ...., n also the sum of number of natural number in each group is equal to the last number of that group i.e., 1 = 1 = last number of first group 1 + 2 = 3 = last number of second group 1 + 2 + 3 = 6 = last number of third group ⇒ 1 + 2 + ..... + n-1= last number of (n-1)th group = (n-1)(n-1+1)/2 = n(n-1)/2 Therefore first number of nth group= n(n-1)/2 + 1 Nth group contains N consecutive natural numbers i.e., with common difference 1 and first term=n(n-1)/2 +1 Thus the sum of numbers in the nth group is = n/2*[ (n[n-1]/2 + 1) + (n-1)*1] Thankyou! Cheers!

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