63. Find the sum of all natural number between 100 and 200 which are d

1.	63. Find the sum of all natural number between 100 and 200 which are divisible by 3.
Answer» o. (a)= 102last no. ( l) = 198difference (d) = 3To find,total no. between 100 and 200 which is DIVISIBLE by 3l= a+(n-1) d198= 102+(n-1)3198-102=(n-1)396=3n-396+3=3n99=3nn=99÷3n=33therefore total no. = 33Now,SUM =n/2(a+l)= 33÷2(102+198)= 33÷2×300=33×100=33000 ANS.

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63. Find the sum of all natural number between 100 and 200 which are divisible by 3.​

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63. Find the sum of all natural number between 100 and 200 which are divisible by 3.