Question

sum of all natural numbers lying between 250 and 1000 which are exactly divisible by 3 is

Answer 1

The required sum is

252+255+....+999

Therefore, a1=252, an=999

so, d=3

To Find n,

WKT, an=a1+(n-1)d

=> n=250

WKT, Sn=(n/2)*(a1+an)

Therefore, S250=(250/2)*(252+999)

=> S250= 156375 (Ans)

Answer 2

The required sum is

252+255+....+999

a1=252, an=999

d=3

To Find n,

an=a1+(n-1)d

n=250

Sn=(n/2)*(a1+an)

S250=(250/2)*(252+999)

S250= 156375 (Ans)