Question

find the sum of all natural numbers lying between 100 and 200 which are divisible by 4.
PLEASE GIVE THE ANSWER STEP BY STEP.
AND PLEASE DON'T POST SILLY ANSWER.

Answer 1

we have to find the sums of 100,104,108,112,.....,200
the above series is in AP
So, first term is 100
difference 104-100=108-104=4
last term = 200
so 200 = 100+(n-1)*4
or 100=(n-1)*4
or 25=n-1
or n=26
now total sum = 13 (2*100+(26-1)*4)=3900

Answer 2

Natural no lying bt 100 and 200 which are divisible by 4 are

104 , 108 , 112 ............................196

We get an ap,

a = 104 , d = 108-104 =4 , An = 196

An = a + (n - 1) d

196 = 104 + (n - 1)4

196 -104 =(n - 1)4

92 / 4 = (n-1)

23 = n-1

n = 23 + 1 =24

Sn = n/2 [2a + (n-1)d]

Sn = 24/2 [ 2(104) + (24-1) 4]

Sn = 12 ( 208 + 23*4)

Sn = 12 (208 + 92)

Sn = 12*300

Sn = 3600