Find the sum of even numbers between 100 and 1000 which are divisible

1.	Find the sum of even numbers between 100 and 1000 which are divisible by 6
Answer» This is a sequence and series based sum. Hence we are using the concept of Arithmetic Progression for solving it. As all numbers divisible by 6 must be divisible by 2. So numbers divisible by 6 are only even numbers. First Term ( a ) = 102 Common Difference ( d ) = 6 Number of terms = ? Last term ( l ) = 996 So Using the formula below, we get, l = a + ( n - 1 ) d => 996 = 102 + ( n - 1 ) 6 => 996 - 102 = 6 ( n - 1 ) => 894 = 6 ( n - 1 ) => 894 / 6 = ( n - 1 ) => 149 = ( n - 1 ) => n = 149 + 1 = 150 terms Hence Sum of all terms can be found out by applying the formula: S ( n ) = n / 2 [ a + l ] => S ( 150 ) = 150 / 2 [ 102 + 996] => S ( 150 ) = 75(1098) => S ( 150 ) = 82350 Hence the sum of all even integers divisible by 6 present between 100 and 1000 is 82350.

Find the sum of even numbers between 100 and 1000 which are divisible by 6

Answer»

This is a sequence and series based sum. Hence we are using the concept of Arithmetic Progression for solving it.

As all numbers divisible by 6 must be divisible by 2. So numbers divisible by 6 are only even numbers.

First Term ( a ) = 102

Common Difference ( d ) = 6

Number of terms = ?

Last term ( l ) = 996

So Using the formula below, we get,

l = a + ( n - 1 ) d

=> 996 = 102 + ( n - 1 ) 6

=> 996 - 102 = 6 ( n - 1 )

=> 894 = 6 ( n - 1 )

=> 894 / 6 = ( n - 1 )

=> 149 = ( n - 1 )

=> n = 149 + 1 = 150 terms

Hence Sum of all terms can be found out by applying the formula:

S ( n ) = n / 2 [ a + l ]

=> S ( 150 ) = 150 / 2 [ 102 + 996]

=> S ( 150 ) = 75(1098)

=> S ( 150 ) = 82350

Hence the sum of all even integers divisible by 6 present between 100 and 1000 is 82350.