A sequence of non-zero numbers is a Geometric Progression (G.P.) if the ratio of any term and its preceding term is always constant.
1, 3, 9, 27, ... is a geometric progression (G.P.) with a = 1 and r = 3
2, 4, 8, 16, ... is a geometric progression (G.P.) with a = 2 and r = 2
nth term of a geometric progression (G.P.)
$MF#%t_n = ar^{n-1}$MF#%
where tn = nth term, a= the first term , r = common ratio, n = number of terms
Example 1 : Find the 10th term in the series 2, 4, 8, 16, ...
a = 2, r = $MF#%\dfrac{4}{2}$MF#% = 2, n = 10
10th term, t10 = $MF#%ar^{n-1} = 2 \times 2^{10-1} = 2 \times 2^{9} = 2 \times 512 = 1024$MF#%
Example 2 : Find 5th term in the series 5, 15, 45, ...
a = 5, r = $MF#%\dfrac{15}{5}$MF#% = 3, n = 5
5th term, t5 = $MF#%ar^{n-1} = 5 \times 3^{5-1}$MF#% = 5 × 34 = 5 × 81 = 405
Sum of first n terms in a geometric progression (G.P.)
$MF#%S_n =
\begin{cases}
\dfrac{a(r^n - 1)}{r - 1}\ & \text{(if r > 1)} \\ \\ \\ \\ \\ \\ \\ \\
\dfrac{a(1 - r^n)}{1 - r}\ & \text{(if r < 1)} \\
\end{cases}$MF#%
where a= the first term , r = common ratio, n = number of terms
Example 1 : Find 4 + 12 + 36 + ... up to 6 terms
a = 4, r = $MF#%\dfrac{12}{4}$MF#% = 3, n = 6
Here r > 1. Hence, $MF#%S_6 = \dfrac{a(r^n - 1)}{r - 1} = \dfrac{4(3^6 - 1)}{3 - 1} = \dfrac{4(729 - 1)}{2} = \dfrac{4 \times 728}{2} = 2 \times 728 = 1456$MF#%
Example 2 : Find $MF#%1 + \dfrac{1}{2} + \dfrac{1}{4} +$MF#% ... up to 5 terms
a = 1, r = $MF#%\dfrac{\left(\dfrac{1}{2}\right)}{1} = \dfrac{1}{2},$MF#% n = 5
Here r < 1. Hence, $MF#%S_6 = \dfrac{a(1 - r^n)}{1 - r} = \dfrac{1\left[1 - \left(\dfrac{1}{2}\right)^5 \right]}{\left(1 - \dfrac{1}{2}\right)}
= \dfrac{\left(1 - \dfrac{1}{32} \right)}{\left(\dfrac{1}{2}\right)} = \dfrac{\left(\dfrac{31}{32}\right) }{\left(\dfrac{1}{2}\right)} = \dfrac{31}{16} = 1\dfrac{15}{16}$MF#%
Sum of an infinite geometric progression (G.P.)
$MF#%S_\infty = \dfrac{a}{1 - r}\ \text{ (if 0 < r < 1)}$MF#%
where a= the first term , r = common ratio
Example$MF#%\text{ : Find }1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +\text{ . . . }\infty$MF#%
a = 1, r = $MF#%\dfrac{\left(\dfrac{1}{2}\right)}{1} = \dfrac{1}{2}$MF#%
Here 0 < r < 1. Hence, $MF#%S_\infty = \dfrac{a}{1 - r} = \dfrac{1}{\left(1 - \dfrac{1}{2}\right)} = \dfrac{1}{\left(\dfrac{1}{2}\right)} = 2$MF#%
Geometric Mean
If three non-zero numbers a, b, c are in G.P., b is the Geometric Mean (G.M.) between a and c. In this case, $MF#% b = \sqrt{ac}$MF#%
The Geometric Mean (G.M.) between two numbers a and b = $MF#%\sqrt{ab}$MF#%
(Note that if a and b are of opposite sign, their G.M. is not defined.)
If a, b, c are in G.P., b2 = ac
If a, b, c are in G.P., $MF#%\dfrac{a-b}{b-c} = \dfrac{a}{b}$MF#%
In a G.P., product of terms equidistant from beginning and end will be constant.
To solve most of the problems related to G.P., the terms of the G.P. can be conveiently taken as
3 terms : $MF#%\dfrac{a}{r}$MF#%, a, ar
5 terms : $MF#%\dfrac{a}{r^2}$MF#%, $MF#%\dfrac{a}{r}$MF#%, a, ar, ar2