1. Tardigrade
  2. Question
  3. Mathematics
  4. Let Sr denotes the sum of the first r terms of an A.P., then (S3r-Sr-1/S2r-S2r-1) is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Let $S_r$ denotes the sum of the first r terms of an A.P., then $\frac{S_{3r}-S_{r-1}}{S_{2r}-S_{2r-1}}$ is equal to

Sequences and Series

Solution:

$\frac{S_{3r} -S_{r-1}}{S_{2r}-S_{2r}-1} $
$\frac{ \frac{3r}{2}\left[2a+\left(3r-1\right)d\right]-\frac{r-1}{2}\left[2a+\left(r-2\right)d\right]}{T_{2r}} $
$ =\frac{ 3ar+\frac{3r\left(3r-1\right)d}{2}-\left(r-1\right)a -\frac{\left(r-1\right)\left(r-2\right)}{2} d}{a+\left(2r-1\right)d} $
$ = \frac{2ar+a+\frac{d}{2}\left[9r^{2}-3r-r^{2}+3r-2\right]}{a+\left(2r-1\right)d}$
$\frac{ a\left(2r+1\right)+\frac{d}{2}\left(8r^{2}-2\right)}{a+\left(2r-1\right)d} $
$ = \frac{a\left(2r+1\right)+d\left(4r^{2}-1\right)}{a+\left(2r-1\right)d}$
$=\frac{\left(2r+1\right)+\left[a+\left(2r-1\right)d\right]}{a+\left(2r-1\right)d} $
$= 2r+1$