Question

A series whose nth term is$\left(\frac{n}{x}\right)+y$, the sum of $r$ terms will be

• A. $\left[\frac{r(r+1)}{2x}\right]+ry$
• B. $\frac{r(r-1)}{2x}$
• C. $\left[\frac{r(r-1)}{2x}\right]–ry$
• D. $\left[\frac{r(r+1)}{2y}\right]+rx$

Answer 1

The correct option is A

$\left[\frac{r(r+1)}{2x}\right]+ry$

Explanation for the correct option:

Step 1. Find the sum of $r$ terms.

given, $a_n=\frac{n}{x}+y$

$\Rightarrow$ $a_1=\frac{1}{x}+y$

$a_2=\frac{2}{x}+y$ and so on

Step 2. Find the Common difference of the given AP.

$\Rightarrow d=a_2-a_1$

$=\frac{2}{x}+y-\frac{1}{x}-y$

$=\frac{1}{x}$

Step 3. Put the value of $a_1$ and $d$ in the formula of summation.

$\Rightarrow {\mathit{S}}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{2}\mathit{r}\mathbf{}\mathbf{[}\mathbf{2}{\mathit{a}}_{\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{(}\mathit{r}\mathbf{-}\mathbf{1}\mathbf{\left)}\mathit{d}\mathbf{\right]}$

$\Rightarrow \frac{r}{2}\left(\frac{2}{x}+2y+(r-1)\frac{1}{x}\right)$

$\Rightarrow \frac{\mathbf{r}\mathbf{(}\mathbf{r}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{2}\mathbf{x}}\mathbf{}\mathbf{+}\mathit{r}\mathit{y}$

Hence, Option ‘A’ is Correct.