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Question
How many terms of the A.P. 27, 24, 21, ..., should be taken so that their sum is zero?
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Solution
A.P. = 27, 24, 21,...
a = 27
d = 24 – 27 = –3
Sn = 0
Let n terms be there in A.P.
Sn = `n/(2)[2a + (n -1)d]`
⇒ 0 = `n/(2)[(2 xx 27) + (n - 1)(-3)]`
⇒ 0 = n[54 – 3n + 3]
⇒ n[57 –3n] = 0
⇒ (57 – 3n) = `(0)/n` = 0
⇒ 3n = 57
∴ n = `(57)/(3)`
= 19
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Q.15
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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