Question

find the sum of all natural numbers between hundred and 200 which are divisible by 3​

Answer 1

To find the sum of the numbers between 100 and 200 which are divisible by 3

Consider a and d to be the first term and common difference respectively

★The AP would be:

102,105,............,198

Here,a=102 and d=3

First we need to find the number of terms

We know that,

l=a+(n-1)d

→198=102+(n-1)3

→96=3n-3

→3n=93

→n=31

Now,

Sum of the terms:

$\sf{\frac{n}{2} (a + l)} \\ \\ = \sf{ \frac{31}{2}(198 + 102) } \\ \\ = \frac{31}{2} \times 300 \\ \\ = 150 \times 31 \\ \\ = \underline{4650}$

The sum of the terms is 4650

Answer 2

Answer:

4950

Step-by-step explanation:

you can get first term and last term easily by simply thinking about the 3 closest digits to 200 and 100(of course natural numbers)