Recent content by serectrus

an=a+(n-1)d 1001=3+(n-1)2 1001-3=(n-1)2 998/2=n-1 499=n-1 499+1=n n=500 500 odd integers from 3-1001 Therefore, 250 2s 250*2=500 500+1=501 Therefore, the sum of the sequence=501 :wink:

an=nth term of an AP (Arithmetic progression) a=1st term d=common difference Since the 1st 3 digit no. divisible by 5 is 100 a=100 the last 3 digit no. divisible by 5 is 995 an=995 common difference (d)=5 n=no. of 3 digit nos. divisible by 5 an=a+(n-1)d 995=100+(n-1)5...