Question

(ii) Find the sum of all natural numbers between 100 and 200 which are divisible
by 4
​

Answer 1

Step-by-step explanation:

this is the answer.

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Answer 2

Answer:

The numbers between 100 and 200 which are divisible by 4 are 104,108,112,...,196.

This is an AP with first term a=104, common difference d=4 and last term l=196.

Let there be n terms in this AP. Then,

a

n

=196⇒a+(n−1)d

196=104+(n−1)×4

∴n=24

∴ Required sum =S

n

=

2

n

[a+l]=

2

24

[104+196]=3600