Let S_n denote the sum of n terms of an A.P. | vedclass

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(B) Given $S_{2n} = 3S_n$. Using the formula for the sum of $n$ terms of an $A.P.$,$S_n = \frac{n}{2}[2a + (n-1)d]$. So,$\frac{2n}{2}[2a + (2n-1)d] =

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$6$

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(B) Given $S_{2n} = 3S_n$. Using the formula for the sum of $n$ terms of an $A.P.$,$S_n = \frac{n}{2}[2a + (n-1)d]$. So,$\frac{2n}{2}[2a + (2n-1)d] = 3 	imes \frac{n}{2}[2a + (n-1)d]$. $2[2a + (2n-1)d] = 3[2a + (n-1)d]$. $4a + 4nd - 2d = 6a + 3nd - 3d$. $nd + d = 2a$,which means $2a = (n+1)d$. Now,we need to find $\frac{S_{3n}}{S_n} = \frac{\frac{3n}{2}[2a + (3n-1)d]}{\frac{n}{2}[2a + (n-1)d]}$. $= 3 	imes \frac{2a + (3n-1)d}{2a + (n-1)d}$. Substitute $2a = (n+1)d$: $= 3 	imes \frac{(n+1)d + (3n-1)d}{(n+1)d + (n-1)d} = 3 	imes \frac{4nd}{2nd} = 3 	imes 2 = 6$.

Let $S_n$ denote the sum of $n$ terms of an $A.P.$ If $S_{2n} = 3S_n$,then the ratio $\frac{S_{3n}}{S_n} = $

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