Let Sn denotes the sum of …
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Sia ? 6 years, 4 months ago
Let 1st term be a and common difference be d.
Given, S2n = 3sn
{tex}\therefore{/tex} {tex}\frac { 2 n } { 2 }{/tex} [2a + (2n - 1) d] = 3 {{tex}\frac { n } { 2 }{/tex} [2a + (n - 1) d]}
{tex}\Rightarrow{/tex} 4a + (4n - 2) d = 6a + (3n - 3)d
{tex}\Rightarrow{/tex} 2a = (n + 1)d ...(i)
Now, {tex}\frac { S _ { 3 n } } { S _ { n } } = \frac { \frac { 3 n } { 2 } [ 2 a + ( 3 n - 1 ) d ] } { \frac { n } { 2 } [ 2 a + ( n - 1 ) d ] }{/tex}
= {tex}\frac { 3 \{ ( n + 1 ) d + ( 3 n - 1 ) d \} } { ( n + 1 ) d + ( n - 1 ) d }{/tex} [from Eq. (i)]
= {tex}\frac { 3 ( 4 n d ) } { 2 n d }{/tex} = 6:1
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