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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, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
Solution
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, tn = 140
But, tn = 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, Sn = `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.
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