Let Sn denote the sum of the first n-terms of an A.P. If S2n=3⋅Sn then S3n÷Sn

S2n=3.Sn⇒2n2[2a+(2n−1)d]=3⋅n2[2a+(n−1)d]⇒4a+22n-1d=6a+3n-1d⇒2a=(n+1)d∴s3nsn=3n22a+(3n−1)dn[2a+(n−1)d2=3(4nd)2nd=6

Let Sn denote the sum of the first n-terms of an A.P. If S2n=3⋅Sn then S3n÷Sn

S2n=3.Sn⇒2n2[2a+(2n−1)d]=3⋅n2[2a+(n−1)d]⇒4a+22n-1d=6a+3n-1d⇒2a=(n+1)d∴s3nsn=3n22a+(3n−1)dn[2a+(n−1)d2=3(4nd)2nd=6

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$Let S_{n} denote the sum of the first n-terms of an A.P. If S_{2 n} = 3 \cdot S_{n} then S_{3 n} \div S_{n}$

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Detailed Solution

$S_{2 n} = 3 . S_{n}$

$\begin{array}{l} \Rightarrow \frac{2 n}{2} [2 a + (2 n - 1) d] = 3 \cdot \frac{n}{2} [2 a + (n - 1) d] \\ \Rightarrow 4 a + 2 (2 n - 1) d = 6 a + 3 (n - 1) d \\ \Rightarrow 2 a = (n + 1) d \\ ∴ \frac{s_{3 n}}{s_{n}} = \frac{\frac{3 n}{2} [2 a + (3 n - 1) d]}{\frac{n [2 a + (n - 1) d}{2}} = \frac{3 (4 n d)}{2 n d} = 6 \end{array}$