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Question
Let Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn then S3n: Sn is equal to ______.
Options
4
6
8
10
Solution
Let Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn then S3n: Sn is equal to 6.
Explanation:
Sn = `n/2 [2a + (n - 1)d]`
∴ S2n = `(2n)/2 [2a + (2n - 1)d]`
3Sn = `(3n)/2 [2a + (3n - 1)d]`
We have S2n = 3 · Sn
⇒ `(2n)/2 [2a + (2n - 1)d] = 3 * n/2 [2a + (n - 1)d]`
⇒ 2[2a + (2n – 1)d] = 3[2a + (n – 1)d]
⇒ 4a + (4n – 2)d = 6a + (3n – 3)d
⇒ 6a + (3n – 3)d – 4a – (4n – 2)d = 0
⇒ 2a + (3n – 3 – 4n + 2)d = 0
⇒ 2a + (– n – 1)d = 0
⇒ 2a – (n + 1)d = 0
⇒ 2a = (n + 1)d ....(i)
Now S3n: Sn = `(3n)/2 [2a + (3n - 1)d] : n/2 [2a + (n - 1)d]`
= `((3n)/2 [2a + (3n - 1)d])/(n/2 [2a + (n - 1)d])`
= `(3[2a + (3n - 1)d])/(2a + (n - 1)d)`
= `(3[(n + 1)d + (3n - 1)d])/((n + 1)d + (n - 1)d)`
= `(3d[n + 1 + 3n - 1])/(d(n + 1 + n - 1))`
= `(3[4n])/(2n)`
= 6
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