Question

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

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

Answer 1

The correct option is A $\left\{\frac{r(r+1)}{2x}\right\}+ry$

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

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

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

$\therefore d=a_2-a_1=\frac{1}{x}$

where, $d$ is common difference of given $AP$.

$\therefore S_r=\frac{r}{2}[2a_1+(r-1\left)d\right]$

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

$=\frac{r(r+1)}{2x}+ry$.