Arithmetic Mean in AP - Formula and Examples

The arithmetic mean is the average(mean) of a number. It is calculated by dividing the sum of all numbers by the number of terms. In real life, we use arithmetic means to get general ideas of the data and also to get the average of wages.

In this article, we will cover the concept of the Arithmetic mean. This category falls under the broader category of Sequence and series, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of ten questions have been asked on this concept, including two in 2014, three in 2017, one in 2019, one in 2020, one in 2021, and two in 2023.

Arithmetic Progression

An arithmetic progression is a sequence in which each term increases or decreases by a constant term or fixed number. This fixed number is called the common difference of an AP and is generally denoted by ' $d$ '.

If $a_1,a_2,a_3,a_4\dots \dots \dots a_{n-1},a_n$ are in AP,

Then $d=a_2-a_1=a_3-a_2=\dots \dots \dots \dots$ = $a_n-a_{n-1}$

In AP, the first term is generally denoted by ‘$a$’

$\mathrm{Eg},1,4,7,10,\dots$ is an AP with a common difference 3
Also, $2,1,0,-1,\dots$ is an AP with a common difference of -1
In AP, the first term is generally denoted by 'a'

General Term of an AP

We found a formula for the general term of a sequence, we can also find a formula for the general term of an arithmetic sequence. First, write the first few terms of a sequence where the first term is 'a' and the common difference is ' $d$ '. We will then look for a pattern.

i.e. $a,a+d,a+2d,a+3d,\dots \dots \dots$

Then the n^th term (general term) of the A.P. is ${\mathit{a}}_{\mathit{n}}\mathit{=}\mathit{a}\mathit{+}\mathit{(}\mathit{n}\mathit{-}\mathit{1}\mathit{)}\mathit{d}$.

$a_1=a+(1-1)d=a$

$a_2=a+(2-1)d=a+d$

$a_3=a+(3-1)d=a+2d$

$a_4=a+(4-1)d=a+3d\dots \dots a_n=a+(n-1)d=l=$ last term

On simplification of the general term, we can see that the general term of an AP is always linear in $n$

$T_n=an+b$

Arithmetic Mean

The sum of all numbers in a list divided by the number of terms in a list is called the Arithmetic Mean.

If three terms are in AP , then the middle term is called the Arithmetic Mean (A.M.) of the other two numbers. So if $a,b$, and $c$ are in A.P., then $b$ is $AM$ of $a$ and $c$.

If $a_1,a_2,a_3,.....,a_n$ are n positive numbers, then the Arithmetic Mean of these numbers is given by 4

$A=\frac{a_1+a_2+a_3+.....+a_n}{n}$

Insertion of $n$-Arithmetic Mean Between $a$ and $b$
If there are n arithmetic mean between two numbers then all the mean are in arithmetic progression (AP).

If ${\mathrm{A}}_1,{\mathrm{A}}_2,{\mathrm{A}}_3....,{\mathrm{A}}_{\mathrm{n}}$ are n arithmetic mean between two numbers a and b, then, $\mathit{a},{\mathrm{A}}_1,{\mathrm{A}}_2,{\mathrm{A}}_3....,{\mathrm{A}}_{\mathrm{n}},\mathit{b}$is an A.P.

Let $d$ be the common difference of this A.P. Clearly, this A.P. contains $n+$ 2 terms.

$\therefore b=(n+2{)}^{\text{th }}$ term

$\Rightarrow b=a+((n+2)-1)d=a+(n+1)d$

$\Rightarrow d=\frac{b-a}{n+1}$

Now, $A_1=a+d=(a+\frac{b-a}{n+1})$

$A_2=a+2d=(a+\frac{2(b-a)}{n+1})\dots \dots A_n=a+nd=(a+n\left(\frac{b-a}{n+1}\right))$

The sum of the n arithmetic mean between two numbers is n times the single A.M. between them.

Proof:

Let ${\mathrm{A}}_1,{\mathrm{A}}_2,{\mathrm{A}}_3....,{\mathrm{A}}_{\mathrm{n}}$ be n arithmetic mean of two numbers a and b.

Then, $\mathit{a},{\mathrm{A}}_1,{\mathrm{A}}_2,{\mathrm{A}}_3....,{\mathrm{A}}_{\mathrm{n}},\mathit{b}$is an A.P. with a common difference$\frac{b-a}{n+1}.$

${\mathrm{A}}_1+{\mathrm{A}}_2+{\mathrm{A}}_3+\dots \dots +{\mathrm{A}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[{\mathrm{A}}_1+{\mathrm{A}}_{\mathrm{n}}]=\frac{\mathrm{n}}{2}[a+b]$

$=\mathrm{n}\left(\frac{a+b}{2}\right)=\mathrm{n}\times (\mathrm{A}.\mathrm{M}$. between $a$ and $b)$

$[\because a,{\mathrm{A}}_1,{\mathrm{A}}_2,\dots ,{\mathrm{A}}_{\mathrm{n}},\mathrm{b}$ is an A.P., $\therefore a+b=A_1+A_n]$

Recommended Video Based on Arithmetic Mean

Solved Examples Based on Arithmetic Mean

Example 1: For three positive integers $\mathrm{p},\mathrm{q},\mathrm{r},x^{p{q}^2}=y^{qr}=z^{p^{2}r}\text{ and }\mathrm{r}=\mathrm{pq}+1$ such that $3,3{\log }_{y}x,3{\log }_z\mathrm{y}\text{,}$ $\text{ 7 }{\log }_{x}z$ are in A.P. with common difference $\frac{1}{2}\text{. Then }r-p-q\text{ is equal to }$ [JEE MAINS 2023]

Solution

Given,

${\mathrm{x}}^{{\mathrm{pq}}^2}={\mathrm{y}}^{\mathrm{qr}}={\mathrm{z}}^{{\mathrm{p}}^2\mathrm{r}}\mathrm{\&}\mathrm{r}=\mathrm{pq}+1$

$3,3{\log }_y^x,3{\log }_z^y,7{\log }_x^z\text{ are in A.P. }$

Now

$\begin{array}{rl}& 3{\log }_y^x=3+\frac{1}{2}=\frac{7}{2}\Rightarrow {\log }_y^x=\frac{7}{6}\\ & x^6=y^7.....(1)\\ & 3{\log }_z^y=3+1=4\Rightarrow {\log }_z^y=\frac{4}{3}\\ & y^3=z^4.....(2)\end{array}$

$\begin{array}{rl}& 7{\log }_x^z=3+\frac{3}{2}=\frac{9}{2}\Rightarrow {\log }_x^z=\frac{19}{14}\\ & z^{14}=x^9.....(3)\end{array}$

Now

$\begin{array}{rl}& x^{{\mathrm{pq}}^2}=x^{\frac{6}{7}q\mathrm{r}}=x^{\frac{9p^2\mathrm{r}}{14}}\\ & {\mathrm{pq}}^2=\frac{6}{7}\mathrm{qr}=\frac{9}{14}{\mathrm{p}}^2\mathrm{r}\\ & \mathrm{pq}=\frac{6}{7}\mathrm{r}{\mathrm{q}}^2=\frac{9}{14}\mathrm{pr}\\ & \mathrm{r}=\mathrm{pq}+1\Rightarrow q^3=\frac{9}{14}\frac{6}{7}\mathrm{r}\cdot r\\ & \Rightarrow r=\frac{6}{7}r+1\\ & \Rightarrow r=7\Rightarrow q=3\end{array}$

Now

$\begin{array}{rl}& \mathrm{r}-\mathrm{p}-\mathrm{q}\\ & =7-2-3\\ & =2\end{array}$

Hence the answer is 2.

Example 2 Let be in an arithmetic progression, with $x_1=2$and their mean equal to 200, If $y_1=i(x_1-i),1⩽i⩽100$then the mean of $y_1,y_2...y_{100}$ is: [JEE MAINS 2023]

Solution

Let the common difference of the AP obtained be $d$.

$\begin{array}{rl}& \therefore \text{ First }AM=a+d\\ & \text{ Last mean }=100-d\\ & \text{ Their ratio }=\frac{a+d}{100-d}=\frac{1}{7}\Rightarrow 7a+8d=100\end{array}$

Also $100=\mathrm{a}+(\mathrm{n}+2-1)\mathrm{d}$

$\begin{array}{rl}& \Rightarrow 100=\mathrm{a}+(\mathrm{n}+1)\mathrm{d}\\ & \Rightarrow 100=33-\mathrm{n}+(\mathrm{n}+1)\mathrm{d}(\text{ given }\mathrm{a}+\mathrm{n}=33)\\ & \Rightarrow \mathrm{d}=\frac{\mathrm{n}+67}{\mathrm{n}+1}\end{array}$

Using (i)

$\begin{array}{rl}& 7(33-n)+\frac{8(\mathrm{n}+67)}{\mathrm{n}+1}=100\\ & \Rightarrow 7{\mathrm{n}}^2-132\mathrm{n}-667=0\\ & \Rightarrow \mathrm{n}=23\end{array}$

Hence, the value of $n=23$.

Example 4 :The common difference of the $A.P.b_1,b_2,\dots b_m$ is 2 more than the common difference of A.P. $a_1,a_2,\dots a_n$. If $a_{40}=-159$, $a_{100}=-399$, and $b_{100}=a_{70}$, then $b_1$ is equal to :
[JEE MAINS 2022]

Solution

Given,

$\begin{array}{rl}& a_{40}=-159\\ & a_{100}=-399\\ & b_{100}=a_{70}\\ & a_1,a_2,\dots ,\text{ an }\to (CD=d)\\ & {\mathrm{b}}_1,{\mathrm{b}}_2,\dots ,{\mathrm{b}}_{\mathrm{m}}\to (\mathrm{CD}=\mathrm{d}+2)\\ & a_{40}=a+39d=-159\\ & a_{100}=a+99d=-399\end{array}$

Subtract : $60d=-240\Rightarrow d=-4$
Using equation (1)

$\begin{array}{rl}& a+39(-4)=-159\\ & a=156-159=-3\\ & a_{70}=a+69d=-3+69(-4)=-279=b_{100}\\ & b_{100}=-279\\ & b_1+99(d+2)=-279\\ & b_1-198=-279\Rightarrow b_1=-81\end{array}$

Hence, the answer is - 81

Example 5: If ${}^n{\mathrm{C}}_4,{}^n{\mathrm{C}}_5$, and ${}^n{\mathrm{C}}_6$ are in A.P., then n can be [JEE MAINS 2020]

Solution:

Arithmetic mean of two numbers (AM)-

$A=\frac{a+b}{2}$

So,

$\begin{array}{rl}& \frac{{}^n{\mathrm{C}}_4+{}^n{\mathrm{C}}_6}{2}={}^n{\mathrm{C}}_5\\ & \frac{n!}{(n-4)!4!}+\frac{n!}{(n-6)!6!}=2\times \frac{n!}{(n-5)!5!}\\ & n=7,14\end{array}$

Hence the value of $n$ can be either 7 or 14 .