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Question

Find the sum of all natural numbers between 1 and 201 which are divisible by 5.

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Solution

Since the sequence is the sum of all natural numbers between 41 and 201, therefore, the sequence is of the type 5+10+15.........+200We know the nth term of an A.P with first term a and the common difference d is:tn=a+(n−1)dHere, the first term is a=5, common difference is d=10−5=5 and tn=200, therefore, tn=a+(n−1)d⇒200=5+(n−1)5⇒200=5+5n−5⇒5n=200⇒n=2005=40We also know the sum of n terms of an A.P with first term a and the common difference d is:Sn=n2[2a+(n−1)d]Here, the first term is a=5, number of terms n=40 and common difference is d=5, therefore, Sn=n2[2a+(n−1)d]=402[(2×5)+(40−1)(5)]=20(10+195)=20×205=4100Hence, the sum of all natural numbers between 41 and 201 is 4100.

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