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.