3.1 Geometric 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:

The geometric mean $(G)$ of two positive numbers $a$ and $b$ is $$G = \sqrt {ab} $$

If $$a,{G_1},{G_2},{G_3},...,{G_n},b$$ are in G.P., then $${G_1},{G_2},{G_3},...,{G_n}$$ are called $n$ geometric means between $a$ and $b$. If $r$ is the common ratio, then $$\begin{equation} \begin{aligned} b = a{r^{n - 1}} \\ \Rightarrow r = {\left( {\frac{b}{a}} \right)^{\frac{1}{{n + 1}}}} \\\end{aligned} \end{equation} $$

$${G_i} = a{r^i} = a{\left( {\frac{b}{a}} \right)^{\frac{i}{{n + 1}}}} = {a^{\frac{{n + 1 - i}}{{n + 1}}}}{b^{\frac{i}{{n + 1}}}}$$ where $i = 1,2,3,...,n$

Note: The product of $n$-geometric means i.e., $${G_1}.{G_2}.{G_3}...{G_n} = {\left( {\sqrt {ab} } \right)^n}$$

Question 13. How many terms of G.P. $3,{3^2},{3^3},...$ are required to give a sum of $120$.

Solution: $a=3$, $r = \frac{{{3^2}}}{3} = 3$ and ${S_n} = 120$

$$\begin{equation} \begin{aligned} 120 = \frac{{a(1 - {r^n})}}{{1 - r}} \\ 120 = \frac{{3(1 - {3^n})}}{{1 - 3}} \\ \frac{{240}}{3} = {3^n} - 1 \\ 81 = {3^n} \\ \therefore n = 4 \\\end{aligned} \end{equation} $$

Question 14. Find the sum of $n$ terms of the series $$11 + 103 + 1005 + ...n{\text{ terms}}$$

Solution: $$\begin{equation} \begin{aligned} 11 + 103 + 1005 + ...n{\text{ terms}} \\ 10 + 1 + 100 + 3 + 1000 + 5 + ...n{\text{ terms}} \\\end{aligned} \end{equation} $$

which shows that the series is a combination of A.P. and G.P.

Considering A.P., $$1 + 3 + 5 + ...\frac{n}{2}{\text{ terms}}$$

$a=1$, $d=3-1=2$

$$\begin{equation} \begin{aligned} {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \\ {\text{ = }}\frac{n}{2}\left[ {2 + \left( {n - 1} \right) \times 2} \right] \\ {\text{ = }}{n^2} \\\end{aligned} \end{equation} $$

Now, considering G.P., $$10 + 100 + 1000 + ...\frac{n}{2}{\text{ terms}}$$

$a=10$, $r = \frac{{100}}{{10}} = 10$

$$\begin{equation} \begin{aligned} {S'}_n = \frac{{a({r^n} - 1)}}{{r - 1}} \\ {\text{ = }}\frac{{10({{10}^n} - 1)}}{{10 - 1}} \\ {\text{ = }}\frac{{10({{10}^n} - 1)}}{9} \\\end{aligned} \end{equation} $$

Therefore, the sum of given series is $${S_n} + {S'}_n = {n^2} + \frac{{10({{10}^n} - 1)}}{9}$$