Question

find the sum of all even numbers between 5 and 100

Answer 1

another approach to solve this rather than sum of A.P

even number between 5 and 100

6,8,10,......,98

since there is written between so we will not include 100


now add 0,2 and 4 to complete series in end we will subtract 0,2 and 4

now series is 0+2+4+6+8+.....+98

now take 2 common

= 2(1+2+3+4+......+49)
sum of n natural number = n(n+1)/2

= 2(49)(50)/2

= 2450

now subtract 2 and 4

= 2450 -2-4 = 2444

thus answer is 2444

Answer 2

Hey there !

Solution:

So the question has asked for sum of all even numbers between 5 and 100.

So 100 will not be included in our case.

So writing the sequence we get,

=> 6, 8, 10, 12, 14 ........, 98

=> a = 6, d = 2, $a_n$ = 98, n = ?

Formula for calculating : $a_n = a + ( n - 1 ) d$

=> 98 = 6 + ( n - 1 ) 2

=> 98 - 6 = ( n - 1 ) 2

=> 92 = ( n - 1 ) 2

=> 92 / 2 = ( n - 1 )

=> 46 = ( n - 1 )

=> n = 46 + 1 = 47

Hence there are 47 terms.

Now, we know that Sum of n terms of an AP can be given as:

$S_n = \frac{n}{2} [ a + a_n ]$

So substituting in the equation we get,

=> Sum of 47 terms = ( 47 / 2 ) ( 6 + 98 )

=> Sum of 47 terms = ( 23.5 ) ( 104 )

=> Sum of 47 terms = 2444

Hence the answer is 2444.

Hope my answer helped !