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

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

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$6$

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(B) Given that ${S_{2n}} = 3{S_n}$.Using the formula for the sum of $n$ terms of an $A.P.$,${S_n} = \frac{n}{2}[2a + (n - 1)d]$.Substituting this into the given equation:$\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$$nd + d = 2a$$2a = (n + 1)d$.Now,we need to find the ratio $\frac{{{S_{3n}}}}{{{S_n}}}$:$\frac{{{S_{3n}}}}{{{S_n}}} = \frac{\frac{3n}{2}[2a + (3n - 1)d]}{\frac{n}{2}[2a + (n - 1)d]} = 3 	imes \frac{2a + (3n - 1)d}{2a + (n - 1)d}$.Substituting $2a = (n + 1)d$ into the expression:$= 3 	imes \frac{(n + 1)d + (3n - 1)d}{(n + 1)d + (n - 1)d} = 3 	imes \frac{(n + 1 + 3n - 1)d}{(n + 1 + n - 1)d} = 3 	imes \frac{4nd}{2nd} = 3 	imes 2 = 6$.

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

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