find the sum of 1 to 100 natura numbers which are divisible by 3​

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1. Answer:

   a=3 d=3 an=99

   Step-by-step explanation:

   an=a+(n-1)d

   99=3+(n-1)3

   99-3=3n-3

   3n=99-3+3

   n=99/3

   n=33

   S33=n/2(2a+32d)

   S33=33/2(2×3+32×3)

   S33=33/2(6+96)

   =33/2×102

   =33×51

   =1683

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