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प्रश्न
First term and the common differences of an A.P. are 6 and 3 respectively; find S27.
Solution: First term = a = 6, common difference = d = 3, S27 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]` - Formula
Sn = `27/2 [12 + (27 - 1)square]`
= `27/2 xx square`
= 27 × 45
S27 = `square`
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उत्तर
It is given that,
First term (a) = 6
Common difference (d) = 3
Sn = `n/2 [bb(2a) + (n - 1)d]` - Formula
∴ S27 = `27/2 [2(6) + (27 - 1)bb((3))]`
= `27/2 (12 + 26 (3))`
= `27/2 (12 + 78)`
= `27/2 xx bb90`
= 1215
Hence, S27 = 1215.
संबंधित प्रश्न
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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`
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Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
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