Question

Find the sum of all natural numbers between 101 and 201 which are divisible by 9?​

Answer 1

Answer:

Since the sequence is the sum of all natural numbers between 41 and 201, therefore, the sequence is of the type 5+10+15.........+200

We know the nth term of an A.P with first term a and the common difference d is:

t n  =a+(n−1)d

Here, the first term is a=5, common difference is d=10−5=5 and t  

n  =200, therefore,  

t  n  =a+(n−1)d

⇒200=5+(n−1)5

⇒200=5+5n−5

⇒5n=200

⇒n=  5 200  =40

We also know the sum of n terms of an A.P with first term a and the common difference d is:

S  n   =  2 n [2a+(n−1)d]

Here, the first term is a=5, number of terms n=40 and common difference is d=5, therefore,  

S  n   =  2 n  [2a+(n−1)d]=  2 40  [(2×5)+(40−1)(5)]=20(10+195)=20×205=4100

Hence, the sum of all natural numbers between 41 and 201 is 4100