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# Sum of n Terms of an AP - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Sum of n terms of an AP is considered one the most difficult concept.

• 46 Questions around this concept.

## Solve by difficulty

Let $\; a_{1},a_{2},a_{3},....$ be terms of an A.P. If $\frac{a_{1}+a_{2}+......+a_{p}}{a_{1}+a_{2}+......+a_{q}}=\frac{p^{2}}{q^{2}},p\neq q,$  then $\; \frac{a_{6}}{a_{21}}\;$ equals:

## Concepts Covered - 1

Sum of n terms of an AP

The sum, Sn  of n terms of an AP with the first term ‘a’ and common difference ‘d’ is given by

$\begin{array}{l}{S_{n}=\frac{n}{2}[2 a+(n-1) d]} \\ {\text { OR }} \\ {S_{n}=\frac{n}{2}[a+l]} \\ {a \rightarrow \text { first term }} \\ {d \rightarrow \text { common difference }} \\ {n \rightarrow \text { number of terms }}\end{array}$

$\\\begin{array}{l}{S_{n}=a+(a+d)+(a+2 d)+\ldots . .+(a+(n-2) d)+(a+(n-1) d) \ldots \ldots .(i)} \\ {\text {Rewrite}} \\ {S_{n}=(a+(n-1) d)+(a+(n-2) d)+\ldots \ldots \ldots+(a-2 d)+(a-d)+a \ldots \ldots(i i)} \\ {(i)+(i i)}\end{array}\\\begin{array}{l}{2 S_{n}=(2 a+(n-1) d)+(2 a+(n-1) d+\ldots \ldots \ldots \ldots . . . . \text { upto } n \text { term }} \\ {S_{n}=\frac{n}{2}[(2 a+(n-1) d)]}\end{array}\\\begin{array}{l}{\text {Also, } a+(n-1) d=l=\text {last term}} \\ {\text {we can also write }} \\ {S_{n}=\frac{n}{2}[a+a+(n-1) d]=\frac{n}{2}[a+l]}\end{array}$

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## Books

### Reference Books

#### Sum of n terms of an AP

Mathematics for Joint Entrance Examination JEE (Advanced) : Algebra

Page No. : 5.7

Line : 1