Let Sn be the sum of the first n terms of an arithmetic progression.

Question

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 :

Accepted Answer

Let a be first term and d be common diff. of this A.P.

Given, S3n = 3S2n

$$ \Rightarrow {{3n} \over 2}[2a + (3n - 1)d] = 3{{2n} \over 2}[2a + (2n - 1)d]$$

$$ \Rightarrow 2a + (3n - 1)d = 4a + (4n - 2)d$$

$$ \Rightarrow 2a + (n - 1)d = 0$$

Now, $${{{S_{4n}}} \over {{S_{2n}}}} = {{{{4n} \over 2}[2a + (4n - 1)d]} \over {{{2n} \over 2}[2a + (2n - 1)d]}} = {{2\left[ {\underbrace {2a + (n - 1)d}_{ = 0} + 3nd} \right]} \over {\left[ {\underbrace {2a + (n - 1)d}_{ = 0} + nd} \right]}}$$

$$ = {{6nd} \over {nd}} = 6$$

Solutions · 1