Question

If $S_r$ denotes the sum of the first $r$ terms of an $AP$ 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. $2r+3$

Answer 1

The correct option is B $2r+1$

If $a$ is the first term of $A.P.$ and $d$ is common difference then

$S_n$: sum of first $n$ terms is

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

Now, $\frac{S_{3r}-S_{r-1}}{S_{2r}-S_{2r-1}}$

$=\frac{\frac{3r}{2}[2a+(3r-1)d]-\frac{r-1}{2}[2a+(r-1-1)d]}{\frac{2r}{2}[2a+(2r-1)d]-\frac{2r-1}{2}[2a+(2r-1-1)d]}$

$=\frac{(\frac{3r}{2}-\frac{r-1}{2})2a+[\frac{3r}{2}(3r-1)-\left(\frac{r-1}{2}\right)(r-2)]d}{(\frac{2r}{2}-\frac{2r-1}{2})2a+[\frac{2r}{2}(2r-1)-\left(\frac{2r-1}{2}(2r-2)\right)]d}$

$=\frac{\left(\frac{2r+1}{2}\right)2a+(4r^2-1)d}{\left(\frac{1}{2}\right)2a+(2r-1)d}$ $=\frac{(2r+1)[\left(a\right)+(2r-1)d]}{\left(a\right)+(2r-1)d}$ $=(2r+1)$