Arithmetic Mean in AP - Formula and Examples

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

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

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

$\mathrm{Eg}, 1,4,7,10, \ldots$ is an AP with a common difference 3
Also, $2,1,0,-1, \ldots$ 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. $a, a+d, a+2 d, a+3 d, \ldots \ldots \ldots$

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

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

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

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

$a_4=a+(4-1) d=a+3 d \ldots \ldots a_n=a+(n-1) d=l=$ 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 $A M$ 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 $\\\mathrm{\;A_1,A_2,A_3....,A_n}$ are n arithmetic mean between two numbers a and b, then, $\mathrm{\mathit{a},A_1,A_2,A_3....,A_n,\mathit{b}}$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, $A_1=a+d=\left(a+\frac{b-a}{n+1}\right)$

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

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

Proof:

Let $\\\mathrm{\;A_1,A_2,A_3....,A_n}$ be n arithmetic mean of two numbers a and b.

Then, $\mathrm{\mathit{a},A_1,A_2,A_3....,A_n,\mathit{b}}$is an A.P. with a common difference$\frac{b-a}{n+1} .$

$\mathrm{A}_1+\mathrm{A}_2+\mathrm{A}_3+\ldots \ldots+\mathrm{A}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{~A}_1+\mathrm{A}_{\mathrm{n}}\right]=\frac{\mathrm{n}}{2}[a+b]$

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

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

Recommended Video Based on Arithmetic Mean

Solved Examples Based on Arithmetic Mean

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

Solution

Given,

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

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

Now

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

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

Now

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

Now

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

Hence the answer is 2.

Example 2 Let be in an arithmetic progression, with $x_1=2$and their mean equal to 200, If $y_1=i(x_1-i),1\leqslant i\leqslant100$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{aligned}
& \therefore \text { First } A M=a+d \\
& \text { Last mean }=100-d \\
& \text { Their ratio }=\frac{a+d}{100-d}=\frac{1}{7} \Rightarrow 7 a+8 d=100
\end{aligned}
$

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

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

Using (i)

$
\begin{aligned}
& 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{aligned}
$

Hence, the value of $n=23$.

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

Solution

Given,

$
\begin{aligned}
& a_{40}=-159 \\
& a_{100}=-399 \\
& b_{100}=a_{70} \\
& a_1, a_2, \ldots, \text { an } \rightarrow(C D=d) \\
& \mathrm{b}_1, \mathrm{~b}_2, \ldots, \mathrm{b}_{\mathrm{m}} \rightarrow(\mathrm{CD}=\mathrm{d}+2) \\
& a_{40}=a+39 d=-159 \\
& a_{100}=a+99 d=-399
\end{aligned}
$

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

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

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{aligned}
& \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{aligned}
$

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