Let S_{n} be the sum of the first n terms of an arithmetic progression. If S_{3n} = 3S_{2n}, then the value of \frac{S_{4n}}{S_{2n}} is: Sequence & Series (Mathematics)

\n\nS_n=\frac{n}{2}(2 n+(n-1) d)\n\n

\n\nthe formula we can mopmers

S_{\text {lend }} S_{\text {ne }}

\n\n

\n\n\begin{gathered}\n\nS_{n n}=\frac{3 n}{2}(2 n+(3 n-1) d) \\\n\nS_{n n}=\frac{2 n}{2}(2 n+(2 n-1) d)=n(2 n+(2 n-1) d)\n\n\end{gathered}\n\n

\n\n3. "Set Up the Given Equation": Accondingts thaprablim, we have:\n\n

\n\n5_m=15_m\n\n

\n\n

\n\n\frac{3 n}{2}(2 n+(3 n-1) d)=3(n(2 n+(2 n-1) d))\n\n

\n\n4. "Simplity tha Equation": Owldingbethsidasby n(maxumingn / ©:\n\n

\n\n\frac{3}{2}(2 n+(3 n-1) d)=3(2 n+(2 n-1) d)\n\n

\n \nNow, multiply both aldou by 2 todelirrinsto the fraction:\n\n

\n\n3\left(2 a+(\operatorname{m} n-1) d^2\right)-6(2 a+(2 n-1) d)\n\n

\n\n5. "Bepard Both Sidea": Equarding both aldes glve:\n\n

\n\n6 x+9 n d-3 d=15 n+6 n d-6 d\n\n

\n\n5. "RearrangingTurra": Rearrangingthesquation:\n\n

\n\n6 a+9 n d-1 d-13 n-6 n d+6 d=0\n\n

\n \nElmpilinging further:\n\n

\n\n-8 n+3 n d+3 d=0\n\n

\n \nDividingthroughtry 3:\n\n

\n\n-5 \mathrm{a}+\mathrm{md}+d=0\n\n

\n \nThus, we hind:\n\n

\n\n2 a=n d+d \quad>\quad 2 n=d(n+1)\n\n

\n\n

\n\nS_{\mathrm{m}}=\frac{4 \mathrm{n}}{2}(\mathrm{~m}+(4 \mathrm{n}-1) \mathrm{d})=\mathrm{mn}(\mathrm{~m} \mathrm{n}+(4 \mathrm{n}-1) \mathrm{d})\n\n

\n\nB. "Qubeituting

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, Subeituta

2 \mathrm{a}=\mathrm{d}(\mathrm{x}+1)

irta

S_{\mathrm{ax}}

\n\n

\n\nS_n=\ln (d(n+1)+(4 n-1) d)=2 n(d(n+1+4 n-1))=\ln (d(5 n))=10 n d^2\n\n

\n\n7. "Ninding S. S. Now, we calculate Shic\n\n

\n\n5 m=n\left(2 n+(2 n-1) d^{\prime}\right)=n(d(n+1)+(2 n-1) d)=n(d(n+1+2 n-1))=n(\n\n

\n \nThandfore, weonhed theratio:\n\n

\n\n\frac{S_m}{S_m}=\frac{10 n d^2}{3 n E^1}=\frac{10}{3}\n\n

\n\n

\n\n\frac{S_{n x}}{S_{n x}}=\frac{10}{3}\n\n

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