Find the sum of all natural numbers between 1 and 145 which are divisible by 4.

Answer

The required sum is 2664.

Explanation

The numbers between 1 and 145 divisible by 4 are 4,8,12,… 144.
Here, `a = t_(1) = 4, d = 4, t_(n) = 144 , n = ? ,S_(n) = ?`
First we find n. brgt `t_(n) = a + ( n -1) d ` … (Formula)
`:. 144 = 4+ ( n 1-) xx 4 ` ….(Substituting the values )
`:. 144 = 4 +4n -4` `:. 4n = 144` `:. n = ( 144)/(4)` `:n = 36`
Now, we find `S_(n) ` i.e., `S_(36)`.
`S_(n) = ( n )/( [t_(1) + t_(n)]`
`= 18 xx 148`
`:. S_(n) = 2664`