Let Sn be the sum of the first n terms of an arithmetic progression. If S3n = 3S2n, then the value of ${{{S_{4n}}} \over {{S_{2n}}}}$ is :
Solution
Let a be first term and d be common diff. of this A.P.<br><br>Given, S<sub>3n</sub> = 3S<sub>2n</sub><br><br>$\Rightarrow {{3n} \over 2}[2a + (3n - 1)d] = 3{{2n} \over 2}[2a + (2n - 1)d]$<br><br>$\Rightarrow 2a + (3n - 1)d = 4a + (4n - 2)d$<br><br>$\Rightarrow 2a + (n - 1)d = 0$<br><br>Now, $${{{S_{4n}}} \over {{S_{2n}}}} = {{{{4n} \over 2}[2a + (4n - 1)d]} \over {{{2n} \over 2}[2a + (2n - 1)d]}} = {{2\left[ {\underbrace {2a + (n - 1)d}_{ = 0} + 3nd} \right]} \over {\left[ {\underbrace {2a + (n - 1)d}_{ = 0} + nd} \right]}}$$<br><br>$= {{6nd} \over {nd}} = 6$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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