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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`.
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Solution
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
∴ \[\boxed{136}\] = 4 + (n – 1) × 4
∴ 136 – 4 = (n – 1) × 4
∴ \[\boxed{132}\] = (n – 1) × 4
∴ `132/4` = n – 1
∴ 33 = n – 1
∴ n = \[\boxed{34}\]
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = `34/2 (4 + 136)`
∴ Sn = 17 × \[\boxed{140}\]
∴ Sn = \[\boxed{2380}\]
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is \[\boxed{2380}\].
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