Question

find out the sum of all natural numbers between 1 and 145 which are divisible by 4​

Answer 1

Answer:

2664

Step-by-step explanation:

Given: To find the sum of all natural numbers between 1 and 145 which are divisible by 4.

The number of natural numbers between 1 and 145 divisible by 4 are

 4, 8, 12, 16, 20....................144

a = 4, d = 4, n = ?,  an = 144

  an = a + (n - 1)d

    144    = 4 + (n - 1)4

   140 = 4n - 4

  4n = 144

 n = 36

Now to find sum we have

Sn = n / 2 [2 a + (n - 1)d]

Sn = 36 / 2 [2(4) + (36 - 1)4]

Sn = 18(8 + 140)

Sn = 18 x 148

Sn = 2664

The sum of natural numbers divisible by 4 between 1 and 145 is 2664.

Hope it was helpful

Answer 2

           

Considering them as in an AP...

4 is the least multiple of 4 after 1.

∴ a_1 = 4

144 is the highest multiple of 4 below 145.

∴ a_n = 144

d = 4

The AP is 4, 8, 12,......, 136, 140, 144.

$n=\frac{a_n-a_1}{d}+1 \\ \\ \\ n=\frac{144-4}{4}+1 \\ \\ \\ n=\frac{140}{4}+1 \\ \\ \\ n=35+1 \\ \\ \\ n=36$

$S_n=\frac{n}{2}[a_1+a_n] \\ \\ \\ S_n=\frac{36}{2}[4+144] \\ \\ \\ S_n=18 \times 148 \\ \\ \\ S_n=\bold{2664}$

∴ 2664 is the answer.

Hope this helps. Plz ask me if you've any doubts.

Thank you. :-))