Question

find the sum of the natural nimbers of between 300 and 600 which are divisible by 6.

Answer 1

If You are including 300 and 600 then 300=a(first tem) and 600 will be the last term...

We know that last term is 600 so...
a + (n-1)d =600 where d=6 and a=300 solving it we get that 600=300+(n-1)6
300=6n-6.........306=6n and n=51
Formula of Sum =n/2(a+l) where l=last term...
so SUM = 51/2(900)=51×400=20400...

Answer 2

hi

aim : Find the sum of the natural nimbers of between 300 and 600 which are divisible by 6.

so the first integer divisible by 6 in 300 to 600

is 300  (300/6 = 50

and last integer is 600 (600/6) = 100

now lets calculate the sum

ans : 300 +306 +..............600

ans : 6 *( 50 +51+...............100)

ans : 6* ( 50 *{ 100 + 50)/2} 

ans : 6 * ( 75*50)

ans :6*3750

ans : 22500

hope it helps u

:)