# 1.4.2 The nth term of a geometric progression

1.4.2 The nth Term of Geometric Progressions

(C) The nth Term of Geometric Progressions

$\overline{)\text{}{T}_{n}=a{r}^{n-1}\text{}}$

a = first term
r = common ratio
n = the number of term
Tn = the nth term

Example 1:
Find the given term for each of the following geometric progressions.
(a) 8 ,4 ,2 ,…… T8
(b) $\frac{16}{27},\text{}\frac{8}{9},\text{}\frac{4}{3},\dots ..$ ,  T6

Solution:
$\begin{array}{l}{T}_{n}=a{r}^{n-1}\\ {T}_{1}=a{r}^{1-1}=a{r}^{0}=a\text{}←\left(\text{First term}\right)\\ {T}_{2}=a{r}^{2-1}=a{r}^{1}=ar\text{}←\left(Second\text{term}\right)\\ {T}_{3}=a{r}^{3-1}=a{r}^{2}\text{}←\left(Third\text{term}\right)\\ {T}_{4}=a{r}^{4-1}=a{r}^{3}\text{}←\left(\text{Fourth term}\right)\end{array}$

(a)
$\begin{array}{l}8,\text{}4,\text{}2,\text{}\dots ..\\ a=8,\text{}r=\frac{4}{8}=\frac{1}{2}\\ {T}_{8}=a{r}^{7}\\ {T}_{8}=8{\left(\frac{1}{2}\right)}^{7}=\frac{1}{16}\end{array}$

(b)
$\begin{array}{l}\frac{16}{27},\text{}\frac{8}{9},\text{}\frac{4}{3},\text{}\dots ..\\ a=\frac{16}{27}\\ r=\frac{{T}_{2}}{{T}_{1}}=\frac{\frac{16}{27}}{\frac{8}{9}}=\frac{2}{3}\\ {T}_{6}=a{r}^{5}\\ =\frac{16}{27}{\left(\frac{2}{3}\right)}^{5}=\frac{512}{6561}\end{array}$

(D) The Number of Term of a Geometric Progression
Smart TIPS: You can find the number of term in an arithmetic progression if you know the last term

Example 2:
Find the number of terms for each of the following geometric progressions.
(a) 2, 4, 8, ….., 8192
(b) $\frac{1}{4},\text{}\frac{1}{6},\text{}\frac{1}{9},\text{}\dots ..,\text{}\frac{16}{729}$
(c) $-\frac{1}{2},\text{1},\text{}-\text{2},.....,\text{64}$

Solution:
(a)
$\begin{array}{l}2,\text{}4,\text{}8,.....,\text{}8192←\left(\text{Last term is given)}\\ a=2\\ r=\frac{{T}_{2}}{{T}_{1}}=\frac{4}{2}=2\\ {T}_{n}=8192\\ a{r}^{n-1}=8192←\left(Then\text{th term of GP,}{T}_{n}=a{r}^{n-1}\right)\\ \left(2\right){\left(2\right)}^{n-1}=8192\\ {2}^{n-1}=4096\\ {2}^{n-1}={2}^{12}\\ n-1=12\\ n=13\end{array}$

(b)
$\begin{array}{l}\frac{1}{4},\text{}\frac{1}{6},\text{}\frac{1}{9},\text{}.....,\text{}\frac{16}{729}\\ a=\frac{1}{4},r=\frac{\frac{1}{6}}{\frac{1}{4}}=\frac{2}{3}\\ {T}_{n}=\frac{16}{729}\\ a{r}^{n-1}=\frac{16}{729}\\ \left(\frac{1}{4}\right){\left(\frac{2}{3}\right)}^{n-1}=\frac{16}{729}\\ {\left(\frac{2}{3}\right)}^{n-1}=\frac{16}{729}×4\\ {\left(\frac{2}{3}\right)}^{n-1}=\frac{64}{729}\\ {\left(\frac{2}{3}\right)}^{n-1}={\left(\frac{2}{3}\right)}^{6}\\ \therefore n-1=6\\ n=7\end{array}$

(c)
$\begin{array}{l}-\frac{1}{2},\text{1},\text{}-\text{2},.....,\text{64}\\ a=-\frac{1}{2},\text{}r=\frac{-2}{1}=-2\\ {T}_{n}=64\\ a{r}^{n-1}=64\\ \left(-\frac{1}{2}\right){\left(-2\right)}^{n-1}=64\\ {\left(-2\right)}^{n-1}=64×-2\\ {\left(-2\right)}^{n-1}=-128\\ {\left(-2\right)}^{n-1}={\left(-2\right)}^{7}\\ n-1=7\\ n=8\end{array}$

(E) Three consecutive terms of a geometric progression
If e, f and g are 3 consecutive terms of GP, then
$\overline{)\text{}\frac{g}{f}=\frac{f}{e}\text{}}$
Example 3:
If p + 20,   p − 4, p −20 are three consecutive terms of a geometric progression, find the value of p.

Solution:
$\begin{array}{l}\frac{p-20}{p-4}=\frac{p-4}{p+20}\\ \left(p+20\right)\left(p-20\right)=\left(p-4\right)\left(p-4\right)\\ {p}^{2}-400={p}^{2}-8p+16\\ 8p=416\\ p=52\end{array}$