QJEEM21Ma0d7b: JEE Main 2021 Mathematics - Class 11 JEEM21Ma0d7b Solution: Let S n be the sum of the first n terms of an arithmetic…

Question

Verified Solution • 2025–26 Session moderate

Let S_n be the sum of the first n terms of an arithmetic progression. If S_3n = 3S_2n, 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, S_3n = 3S_2n

\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

Step-by-Step Verified Solution