Question

Find the sum of all three digit natural numbers which are divisible by 5

Answer 1

Answer:98,550

Step-by-step explanation:A=(100,105,110,.........995)

a=t1=100

Tn=995

d=5

We know that to find n

Tn=a+(n-1)d

995=100+(n-1)5

995=100+5n-5

995=95+5n

995-95=5n

900/5=n

n=180

.'. To find sum of all three digit natural number divisible by 5

We know,

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

S180=180/2(2*100+(180-1)5)

S180=90(200+179*5)

S180=90(200+895)

S180=90*1095

S180=98,550