For a series whose $n^{th}$ term is $\left( \frac{n}{x} \right) + y$,the sum of $r$ terms is:

  • 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
    $\left\{ \frac{r(r + 1)}{2y} \right\} - rx$

Explore More

Similar Questions

Find the sum of the sequence $7, 77, 777, 7777, \ldots$ to $n$ terms.

The sum of $n$ terms of the following series $1 + (1 + x) + (1 + x + x^2) + \dots$ will be

If the sum of the series $\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^2}-\frac{1}{2 \cdot 3}+\frac{1}{3^2}\right)+\left(\frac{1}{2^3}-\frac{1}{2^2 \cdot 3}+\frac{1}{2 \cdot 3^2}-\frac{1}{3^3}\right)+\left(\frac{1}{2^4}-\frac{1}{2^3 \cdot 3}+\frac{1}{2^2 \cdot 3^2}-\frac{1}{2 \cdot 3^3}+\frac{1}{3^4}\right)+\ldots$ is $\frac{\alpha}{\beta}$,where $\alpha$ and $\beta$ are co-prime,then $\alpha+3\beta$ is equal to....

If $S_n = 1^3 + 2^3 + \ldots + n^3$ and $T_n = 1 + 2 + \ldots + n$,then

If $\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i {\sum\limits_{k = 1}^j {1 = 560} } } $,then the value of $n$ is:

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo