Question

Find the sum of all natural numbers between 100 and 400 which are exactly divisible by 4 or 5?

Answer 1

in ap
a1=104
an= 396
n=?
396=104+(n-1)4
396-104=4(n-1)
292/4=n-1
73+1=n
n=74
Now,
Sn = [(n/2)2a+(n-)d]
=(74/2)×208+292
=18500




Hope it's help u !

Answer 2

NUMBERS

A number divisible by both 4 and 5 should be divisible by 20 which is the LCM of 4 and 5.

To find the sum of numbers divisible by 4 or 5 we need to add the sum of numbers divisible by 4 and sum of numbers divisible by 5 then subtract sum of numbers divisible by both 4 and 5 from it.

I am including numbers 100 and 400 both.

Numbers divisible by 4 in the given range$=100,104,108, 112,\_,\_,\_,400$

Sum of numbers divisible by 4 = $100+104+108+112,\_+\_+\_+400$

$\implies 4(25+26+27+\_\_+100)=4(\frac{76}{2})(25+100)=19000$

Sum of numbers divisible by 5 = $100+105+110+115,\_+\_+\_+400=$

$\implies 5(20+21+22+\_\_+80)=5(\frac{61}{2})(20+80)=15250$

Sum of numbers divisible by 20 =$100+120+140+\_+\_,\_,400$

$\implies 20(5+6+7+\_,\_+20)=20(\frac{16}{2})\times(5+20)=4000$

Hence required sum will be $19000+15250-4000=30250$.

So, required sum of all natural number is 30250.