2.1 Arithmetic mean

2.1 Arithmetic mean

3.1 Geometric mean

4.1 Harmonic mean

5.1 Recognization of A.P., G.P. and H.P.

6.1 Important results:

If $a$, $b$ and $c$ are in A.P. then, $$\begin{equation} \begin{aligned} b - a = c - b \\ 2b = a + c \\ b = \frac{{a + c}}{2} \\\end{aligned} \end{equation} $$

So, $b$ is the single arithmetic mean between $a$ and $c$.

${A_1},{A_2},{A_3},...,{A_n}$ numbers are inserted between $a$ and $b$ such that it forms an A.P. in which $$\begin{equation} \begin{aligned} {A_1} \to I{\text{ }}AM \\ {A_2} \to II{\text{ }}AM \\ {A_3} \to III{\text{ }}AM \\ {A_n} \to {n^{th}}{\text{ }}AM \\\end{aligned} \end{equation} $$

So total terms in A.P. is $n+2$. First term is $a$ and last term is $b$.

Therefore,

$$\begin{equation} \begin{aligned} {T_{n + 2}} = a + (n + 2 - 1)d \\ b = a + (n + 1)d \\ d = \frac{{b - a}}{{n + 1}} \\\end{aligned} \end{equation} $$

where, $d$ is the common difference of the A.P. and $n$ is the number of arithmetic mean added.

So, first arithmetic mean is the second term of A.P. i.e., $$a + d = a + \frac{{b - a}}{{n + 1}}$$

Similarly, second arithmetic mean is the third term of A.P. i.e., $$a + 2d = a + 2\left( {\frac{{b - a}}{{n + 1}}} \right)$$

Similarly, $n$th arithmetic mean is $$a + nd = a + n\left( {\frac{{b - a}}{{n + 1}}} \right)$$

Question 9. If $n$ arithmetic mean are inserted between $20$ and $80$ such that $\frac{{{A_1}}}{{{A_n}}} = \frac{1}{3}$. Find $n$.

Solution: $$20,{a_1},{a_2},...,{a_n},80$$ is in A.P. in which first term $a=20$ and last term $b=80$. Therefore, common difference is $$d = \frac{{b - a}}{{n + 1}} = \frac{{80 - 20}}{{n + 1}} = \frac{{60}}{{n + 1}}$$

It is given that

$$\begin{equation} \begin{aligned} \frac{{{A_1}}}{{{A_n}}} = \frac{1}{3} \\ \Rightarrow \frac{{a + d}}{{a + nd}} = \frac{1}{3} \\ \Rightarrow \frac{{20 + \frac{{60}}{{n + 1}}}}{{20 + \frac{{60n}}{{n + 1}}}} = \frac{1}{3} \\ 60n + 240 = 80n + 20 \\ n = 11 \\\end{aligned} \end{equation} $$

Question 10. Prove that the sum of $n$ arithmetic mean i.e., $${A_1} + {A_2} + {A_3} + ... + {A_n} = n\left( {\frac{{a + b}}{2}} \right)$$

Solution: $$\begin{equation} \begin{aligned} LHS \\ {A_1} + {A_2} + {A_3} + ... + {A_n} = \frac{n}{2}\left( {{A_1} + {A_n}} \right) \\ {\text{ = }}\frac{n}{2}\left( {a + d + a + nd} \right) \\ {\text{ = }}\frac{n}{2}\left( {2a + (n + 1)d} \right) \\ {\text{ = }}\frac{n}{2}\left( {2a + (n + 1)\left( {\frac{{b - a}}{{n + 1}}} \right)} \right) \\ {\text{ = }}\frac{n}{2}\left( {2a + b - a} \right) \\ {\text{ = }}n\left( {\frac{{a + b}}{2}} \right) \\ {\text{ = }}RHS \\\end{aligned} \end{equation} $$

Question 11. Between two numbers whose sum is $\frac{{13}}{6}$, even number of arithmetic means are inserted sum of which exceeds their number by unity. How many arithmetic means are inserted?

Solution: Let $n$ arithmetic means are inserted between the two numbers $a$ and $b$ such that $$a + b = \frac{{13}}{6}$$

Sum of $n$ arithmetic means $=\ n+1$

$$\begin{equation} \begin{aligned} n\left( {\frac{{a + b}}{2}} \right) = n + 1 \\ n\left( {\frac{{13}}{{12}}} \right) = n + 1 \\ 13n = 12n + 12 \\ n = 12 \\\end{aligned} \end{equation} $$

Question 12. If $x$ and $y$ are two real numbers such that ${r^{th}}$ mean between $x$ and $2y$ is equal to ${r^{th}}$ mean between $2x$ and $y$ when $n$ AM are inserted between them in both the case. Prove that $$\frac{{n + 1}}{r} - \frac{y}{x} = 1$$

Solution: Case I

$$x,{a_1},{a_2},{a_3},...,{a_n},2y$$

${r^{th}}$ term is $${a_r} = x + rd = x + r\left( {\frac{{2y - x}}{{n + 1}}} \right)$$

Case II

$$2x,{b_1},{b_2},{b_3},...,{b_n},y$$

${r^{th}}$ term is $${b_r} = 2x + rd = 2x + r\left( {\frac{{y - 2x}}{{n + 1}}} \right)$$

According to question, $$\begin{equation} \begin{aligned} {a_r} = {b_r} \\ x + r\left( {\frac{{2y - x}}{{n + 1}}} \right) = 2x + r\left( {\frac{{y - 2x}}{{n + 1}}} \right) \\ r\left( {\frac{{2y - x}}{{n + 1}}} \right) - r\left( {\frac{{y - 2x}}{{n + 1}}} \right) = x \\ \frac{r}{{n + 1}}\left( {2y - x - y + 2x} \right) = x \\ \frac{r}{{n + 1}}\left( {x + y} \right) = x \\ x + y = \frac{{(n + 1)x}}{r} \\ x\left( {1 + \frac{y}{x}} \right) = \frac{{(n + 1)x}}{r} \\ \frac{{n + 1}}{r} - \frac{y}{x} = 1 \\\end{aligned} \end{equation} $$