Question

If $S_r$ denotes the sum of the first r terms of an A.P. Then, $\frac{S_{3r}}{S_{2r}-S_r}$ is ?

• A. $n$
• B. $3n$
• C. $3$
• D. None of these

Answer 1

The correct option is C $3$

Let 'a' be the first term and 'd' be the common difference of an A.P.

Since $S_r$ is the sum of the first r terms of an A.P therefore,

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

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

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

Thus consider,

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

$S_{2r}-S_r=\frac{r}{2}\{[4a+(4r-2)d]-[2a+(r-1)d]\}$

$S_{2r}-S_r=\frac{r}{2}\{4a-2a+4rd-2d-rd+d\}$

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

$S_{2r}-S_r=\frac{1}{3}\times \frac{3r}{2}\{2a+(3r-1)d\}$

$S_{2r}-S_r=\frac{1}{3}\times S_{3r}$

$\Rightarrow$$\frac{S_{3r}}{S_{2r}-S_r}=3$

Hence option C is correct.