Question

Find the sum of natural numbers between 1 to 140, which are divisible by 4.

Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136

Here d = 4, therefore this sequence is an A.P.

a = 4, d = 4, t_n = 136, S_n = ?

t_n = a + (n – 1)d

`square` = 4 + (n – 1) × 4

`square` = (n – 1) × 4

n = `square`

Now,

S_n = `"n"/2["a" + "t"_"n"]`

S_n = 17 × `square`

S_n = `square`

Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.

Answer 1

The numbers from 1 to 140 which are divisible by 4 are 4, 8, 12, ... 140

This sequence is an A.P. with a = 4, d = 8 – 4 = 4, t_n = 140

But, t_n = a + (n − 1)d

∴ 140 = 4 + (n − 1)4

∴ 140 - 4 = (n − 1)4

∴ 136 = (n − 1)4

∴ `136/4` = n − 1 

∴ 34 + 1 = n

∴ n = 35

Now, S_n = `n/2` [2a + (n − 1)d]

∴ S₃₅ = `35/2`[2 × 4 + (35 − 1)4]

= `35/2` [8 + (34)4]

= `35/2`[8 + 136]

= `35/2` × 144

= 35 × 72

∴ S₃₅ = 2520

∴ The sum of all numbers from 1 to 140 which are divisible by 4 is 2520.