Question

find the sum of all natural numbers between 200 and 300 which are divisible both by;-
1). 3 and 4
2). 3 or 4

Answer 1

Answer:

12249

Step-by-step:

From the properties of AP :

Sum of first n terms is ( n / 2 )( a + l )

nth term = a + ( n - 1 )d where a is the first, d is the common difference and l is the last term

Condition 1 : numbers are divisible by 3 and 4, which means numbers are divisible by 3 x 4 ( = 12 ).

First number divisible by 12( after 200 ) is 204

Last number divisible by 12( before 300 ) is 288.

Numbers are increasing by 12, so common difference is 12.

Let the number of terms be n.

= > 288 = 204 + ( n - 1 )12

= > 288 - 204 = ( n - 1 ) 12

= > 8 = n

Thus,

Sum of terms = (8/2)( 204 + 288 ) = 1968.

Situation 2 :

Numbers is either divisible by 3 or 4.

If we say numbers are divisible by 12, then we are wrong. They divisible by either 3 or by 4.

Sum of those terms will be : sum of numbers divisible by 3 + sum of numbers divisible by 4 - sum of numbers divisible by 12.

First numbers divisible by 3 and 4 are 201 and 204 and last numbers are divisible by 3 and 4 are 297 and 296 respectively.

Number of terms are : 33 and 24.

Required sum is :

= > (33/2)(201+297) + (24/2)(204+296) - 1968

= > 8217 + 6000 - 1968

= > 12249