Question

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

• A. 2r - 1
• B. 2r + 1
• C. 4r + 1
• D. 2r + 2

Answer 1

Answer: (B)

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