Arithmetic Means : Problems and Solutions

Definition: Arithmetic Mean

In a finite arithmetic progression, the terms between the first term and the last term are called the arithmetic means.

အဆုံးရှိ A.P ကိန်းစဉ် တစ်ခု၏ ရှေ့ဆုံးကိန်းနှင့် နောက်ဆုံးကိန်းကြားရှိ ကိန်းများအားလုံးကို arithmetic means ဟုခေါ်သည်။

If $\boldsymbol{u}_{1}, \boldsymbol{u}_{2}, \boldsymbol{u}_{3}, \ldots \boldsymbol{u}_{n-1}, \boldsymbol{u}_{n}$ is an A.P., then $\boldsymbol{u}_{2}, \boldsymbol{u}_{3}, \ldots, \boldsymbol{u}_{n-1}$ are called arithmetic means. The arithmetic mean between two numbers $x$ and $y$ is given by

$\begin{array}{|l|}\hline \text {A.M} = \dfrac{x+y}{2}\\ \hline \end{array}$

Exercises

1. Find the A.M. between
   (a) $-3$ and 3.
   (b) $2-\sqrt{2}$ and $2+\sqrt{2}$.
   (c) $\log 3$ and $\log 12$.





$\begin{aligned} &\text{The A.M. between} x \text{ and } y= \frac{x+y}{2} \\\\ \text{(a) } &\text{The A.M. between } -3 \text{ and } 3 \\\\ =&\frac{-3+3}{2} \\\\ =& 0 \\\\ \text{(b) } &\text{The A.M. between } 2-\sqrt{2} \text{ and } 2+\sqrt{2}\\\\ =&\frac{2-\sqrt{2}+2+\sqrt{2}}{2} \\\\ =& 2 \\\\ \text{(c) } &\text{The A.M. between } \log 3 \text{ and } \log 12\\\\ =&\frac{\log 3+\log 12}{2} \\\\ =& 2 \\\\ =&\frac{1}{2} \log 36 \\\\ =&\log \sqrt{36} \\\\ =&\log 6 \end{aligned}$



2. Insert three arithmetic means between $-5$ and $19$ .





Let the required arithmetic means be $x_{1}, x_{2}$ and $x_{3}$. $\begin{aligned} \therefore\ -5, x_{1},\ & x_{2},\ x_{3},\ 19 \text { is an A.P. } \\\\ \therefore\ a &=-5 \\\\ u_{5}&=19 \\\\ a+4 d&=19 \\\\ -5+4 d&=19 \\\\ 4 d&=24\\\\ d&=6 \\\\ x_{1}&=a+d=1 \\\\ x_{2}&=a+2 d=7 \\\\ x_{3}&=a+3 d=13 \end{aligned}$



3. Insert five arithmetic means between $p+q$ and $19 p-11 q$.





Let the required arithmetic means be $x_{1}, x_{2}, x_{3}, x_{4}$ and $x_{5}$. $\therefore p+q, x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, 19 p-11 q$ is an A.P. Let the first term be $a$ and the common difference be $d$. $\begin{aligned} \therefore\ a &=p+q \\\\ u_{7} &=19 p-11 q \\\\ a+6 d &=19 p-11 q\\\\ \therefore\ 6 d &=18 p-12 q \\\\ d &=3 p-2 q \\\\ \therefore\ x_{1} &=a+d=4 p-q \\\\ x_{2} &=a+2 d=7 p-3 q \\\\ x_{3} &=a+3 d=10 p-5 q \\\\ x_{4} &=a+4 d=13 p-7 q \\\\ x_{5} &=a+5 d=16 p-9 q \end{aligned}$



4. If five arithmetic means are inserted between $-10$ and $116$, what is the third A.M.?





Let the fine A.Ms between $-10$ and $n_{6}$ be $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$. $\therefore-10, x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, 116$ is an A.P. Let the first term be $a$ and the common difference be $d$. $\begin{aligned} \therefore\ a &=-10 \\\\ u_{7} &=116 \\\\ a+6 d &=116 \\\\ \therefore\ 6 d &=126 \\\\ d &=21 \\\\ \therefore\ x_{3} &=a+3 d \\\\ &=53 \end{aligned}$



5. If $n$ arithmetic means are inserted between $a$ and $b$, show that the common difference of the A.P. is $\dfrac{b-a}{n+1}$.





Let the $n$ arithmetic means between $a$ and $b$ be $x_{1}, x_{2}, x_{3}, \ldots, x_{n} .$ $\therefore\ a, x_{1}, x_{2}, x_{3}, \ldots, x_{n}, b$ is an A.P. Let the common difference be $d$. $\begin{aligned} u_{n+2} &=b \\\\ a+(n+2-1) d &=b \\\\ (n+1) d &=b-a \\\\ \therefore\ d &=\dfrac{b-a}{n+1} \end{aligned}$



6. If $n$ arithmetic means are inserted between $20$ and $80$ such that the ratio of first mean to the last mean is $1: 3$, find the value of $n$.





Let the $n$ arithmetic means between 20 and 80 be $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$. $\therefore 20, x_{1}, x_{2}, x_{3}, \ldots, x_{n}, 80$ is an A.P. Let the first termbe $a$ and the common differenee be $d$. $\begin{aligned} \therefore \ a &=20 \\\\ u_{n+2} &=80\\\\ a+(n+2-1) d &=80 \\\\ 20+(n+1) d &=80 \\\\ d &=\dfrac{60}{n+1} \\\\ \dfrac{x_{1}}{x_{n}} &=\dfrac{1}{3} \\\\ \dfrac{a+d}{a+n d} &=\dfrac{1}{3}\\\\ 3 a+3 d &=a+n d \\\\ 2 a &=(n-3) d \\\\ 40 &=(n-3) \dfrac{60}{n+1} \\\\ 2 n+2 &=3 n-9 \\\\ n &=11 \end{aligned}$



7. If the A.M. between $p^{\text {th }}$ and $q^{\text {th }}$ terms of an A.P. be equal to the A.M. between $r^{\text {th }}$ and $s^{\text {th }}$ terms of the A.P., show that $p+q=r+s$.





$\begin{aligned} \text { A.M. between } u_{p} \text { and } u_{q} &=\dfrac{u_{p}+u_{q}}{2} \\\\ &=\dfrac{a+(p-1) d+a+(q-1) d}{2} \\\\ &=\dfrac{2 a+(p+q-2) d}{2} \\\\ &=a+\dfrac{1}{2}(p+q-2) d \\\\ \text { A.M. between } u_{r} \text { and } u_{s} &=\dfrac{u_{r}+u_{s}}{2} \\\\ &=\dfrac{a+(r-1) d+a+(s-1) d}{2} \\\\ &=\dfrac{2 a+(r+s-2) d}{2} \\\\ &=a+\dfrac{1}{2}(r+s-2) d\\\\ \text {By the problem}, \quad \quad \quad&\\\\ a+\dfrac{1}{2}(p+q-2) d &=a+\dfrac{1}{2}(r+s-2) d \\\\ \therefore p+q &=r+8 \end{aligned}$



8. If $x, y, z$ are in A.P. and $A_{1}$ is the A.M. between $x$ and $y$, and $A_{2}$ is the A.M. between $y$ and $z$, prove that the A.M. between $A_{1}$ and $A_{2}$ is $y$.





$\begin{aligned} x, y, z \text{ are in A.P}&\\\\ \therefore\quad y&=\dfrac{x+z}{2}\\\\ \text{ A.M between } x \text{ and } y&=\dfrac{x+y}{2}\\\\ \therefore\quad A_{1}&=\dfrac{x+y}{2}\\\\ \text{ A.M between } y \text{ and } z&=\dfrac{y+z}{2}\\\\ \therefore\quad A_{2}&=\dfrac{y+z}{2}\\\\ \text{ A.M between } A_{1} \text{ and } A_{2}&=\dfrac{A_{1}+A_{2}}{2}\\\\ &=\dfrac{1}{2}\left(\dfrac{x+y}{2}+\dfrac{y+z}{2}\right) \\\\ &=\dfrac{1}{2}\left(\dfrac{x+z}{2}+y\right) \\\\ &=\dfrac{1}{2}(y+y) \\\\ &=\dfrac{1}{2}(2 y) \\\\ &=y \end{aligned}$



9. If $x$ is the A.M. between $a$ and $b$, show that $\dfrac{x+2 a}{x-b}+\dfrac{x+2 b}{x-a}=4$.





$\begin{aligned} x \text{ is the }& \text{ A.M. between } a \text{ and } b.\\\\ \therefore \quad x&=\dfrac{a+b}{2}\quad\quad \\\\ \dfrac{x+2 a}{x-b}+\dfrac{x+2 b}{x-a} &=\dfrac{\dfrac{a+b}{2}+2 a}{\dfrac{a+b}{2}-b}+\dfrac{\dfrac{a+b}{2}+2 b}{\dfrac{a+b}{2}-a}\\\\ &=\dfrac{5 a+b}{a-b}+\dfrac{a+5 b}{b-a} \\\\ &=\dfrac{5 a+b}{a-b}+\dfrac{-a-5 b}{a-b} \\\\ &=\dfrac{4 a-4 b}{a-b} \\\\ &=\dfrac{4(a-b)}{a-b}\\\\ &=4 \end{aligned}$

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