Question

find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 3

Answer 1

a=252,d=3,an=999
an=a+(n-1)d
999=252+(n-1)3
999=252+3n-3
999=249+3n
999-249=3n
750=3n
n=750÷3
n=250

Answer 2

Answer:

first no = 252

last number is 999 which divisible by 3

total no be n

an = a+ (n-1)d

999=252+(n-1)3

n=250

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

250/2 * (2*252 + (250-1) *3 )

solve

ans : 156375