Question

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

Answer 1

first no. (a)= 102
last no. ( l) = 198
difference (d) = 3
To find,
total no. between 100 and 200 which is divisible by 3
l= a+(n-1) d
198= 102+(n-1)3
198-102=(n-1)3
96=3n-3
96+3=3n
99=3n
n=99÷3
n=33
therefore total no. = 33
Now,
Sum =n/2(a+l)
= 33÷2(102+198)
= 33÷2×300
=33×100
=33000 Ans.

Answer 2

Answer:

The “sum” of all the multiples of 3 between the numbers “100 and 200” are 4950. Let the no. of terms be denoted as n. Hence, the “sum” of all the multiples of 3 between the numbers “100 and 200” are 4950.