Question

Find the sum of all numbers divisible by 6 between 100 and 400

Answer 1

Hey,sup!

As per the question,

a=102.
d=6.
l=396.

=> T(n)= a+(n-1)d.
=> 396=102+(n-1)6.
=> 396-102=(n-1)6.
=> 294=(n-1)6.
=> n-1= 294/6.
=> n= 49+1.
=> n=50.

=> S(n)= n/2 (a+l).
=> S(50)=50/2(102+396).
= 25×498.
= 12450.

So, the sum of all numbers divisible by 6 between 100 and 400 is 12450.

Hope it helps.

Answer 2

The sum of all numbers divisible by 6 between 100 and 400 is 12450.

Given :

Numbers between 100 and 400.

To find :

The sum of all numbers divisible by 6 between 100 and 400

Solution:

The sequence will be 102,108,114,120,.........,396

Here 1st term a=102(which is the 1st term greater than 100 that is divisible by 6)

The last term less than 400 which is divisible by 6 is 396

The number of terms in the A.P 102,108,114,...396 is given by

$(396-102/6)+1=50$

​Common difference =d=6

So,

$S=25(102+396)\\S=12450$

Therefore, the sum of all numbers divisible by 6 between 100 and 400 is 12450.

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