Sequences & Series Question - None

Question

Given, common difference > 0 and

$2 S_{3n} = S_{4n} - S_{3n}$, where $S_n = Pn^2 + Qn$

Find the ratio $\frac{S_{2n}}{S_{4n} - S_{2n}}$

Accepted Answer

Step 1: From $2 S_{3n} = S_{4n} - S_{3n}$, we get $2S_{3n} = S_{4n} - S_{3n}$

Step 2: Substituting $S_n = Pn^2 + Qn$:
$2[P(3n)^2 + Q(3n)] = P(4n)^2 + Q(4n) - [P(3n)^2 + Q(3n)]$

Step 3: Simplifying: $2Pn^2 + 2Qn = 0$, which gives $Q = -nP$ ... (i)

Step 4: Now, $\frac{S_{2n}}{S_{4n} - S_{2n}} = \frac{P(2n)^2 + Q(2n)}{[P(4n)^2 + Q(4n)] - [P(2n)^2 + Q(2n)]}$

Step 5: $= \frac{2n(2nP + Q)}{12Pn^2 + 2nQ} = \frac{2nP - nP}{6nP - nP} = \frac{2nP}{5nP} = \frac{1}{5}$

∴ Answer is $\frac{1}{5}$

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