The sum of the first $n$ terms of an $A.P.$ is given by $S_{n} = \ldots \ldots \ldots \ldots$
- A$n[2a + (n - 1)d]$
- B$\frac{1}{2} n[2a + (n - 1)d]$
- C$\frac{1}{2} n[a + (n - 1)d]$
- D$\frac{1}{2} n[2a + (n + 1)d]$
Explore More
Similar Questions
For the finite $A.P.$ $1, 4, 7, \ldots, 118,$ find the $15^{th}$ term from the end.
Easy
View SolutionThe digits of a three-digit number form an $A.P.$ and their sum is $15$. The number obtained by reversing the order of the digits exceeds the original number by $594$. Find the original number.
Difficult
View SolutionFor a given $A.P.$,the first term is $-4$ and the common difference is $-5$. Then,the $12^{th}$ term of the $A.P.$ is $\ldots \ldots \ldots$.
Easy
View SolutionWhich term of the $A.P.$ $63, 65, 67, \ldots$ and $3, 10, 17, \ldots$ are equal (in $^{th}$)?
Medium
View SolutionThe $11^{\text{th}}$ term of the $AP: -5, \frac{-5}{2}, 0, \frac{5}{2}, \dots$ is
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo