Question

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

Answer 1

Answer:

natural number are =104,108,112,116,120,124,128,132,136,140,144,148,152,156,160,164,168,172,176,180,184,188,192,196,200

the sum is all natural numbers are

3800

Step-by-step explanation:

to

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