Question

compute the sum of all integers between 100 to 800 that are divisible by 3.

Answer 1

 Here is the method I would use. 
The first integer in this group is 102, and 1/3 of that is 34. 
The last integer in the group is 798, and 1/3 of that is 266. 
So you want to add all the integers from 34 to 266 and triple that sum. 
The sum of the integers from 1 to N is N(N+1)/2. 
Use this formula to calculate the sum of the integers from 1 to 266, then subtract from that, the sum of the integers from 1 to 33. Then don't forget to triple the result.


I hope it will help u my friend

Answer 2

hi there

let us consider this problem as an ap(Arthmetic Progression)

let's start,

the first number which is divisible by 3 between 100 and 800 is 102

therefore, a=102

then,d(the common difference) =3

an(the last term)= 789

now an=a+(n-1)d

> 789=102+(n-1)3

> 3n-3+102=789

> 3n+99=789

> 3n=789-99

> 3n = 690

> n= 690/3

> n=230

therefore the number of terms is 230        

now sn(sum of terms)= n/2{2a+(n-1)d}

= 230/2{2*102+(230-1)3}

= 115{204+229*3}

=115{204+687}

=115*891

=102465

therefore the sum is 102465