Question

find the sum of all numbers from 1 to 140 ehich are divisible by 4

Answer 1

An = a + (n-1 ) d
136 = 4 + (n -1 ) 4
132/4= n-1
33+1 =n
n = 34


Sn = n/2 ( a + l )
= 34/2 ( 4 + 136)
= 17 (140 )
= 2380

Answer 2

Heya User,

--> First multiple of '4' = 4*1
--> Last considerable multiple = 140 = 4*35

--> The sum = [ 4 * 1 + 4 * 2 + 4 * 3 + ... + 4 * 35 ]
                      = 4 * [ 1 + 2 + 3 + ... + 35 ] --> ( i )

--> Easy way out --> 

--> [ 1 + 2 + ... + n ] = { n ( n + 1 ) / 2 }

Putting this in ( i ) -->
--> The sum is --> 4 * 35 * 36 / 2 = 2520 ... 0_0