3. The nth Term of Geometric Progressions (Part 1)

Posted on April 22, 2020 by user

1.4.2 The nth Term of Geometric Progressions

(C) The nth Term of Geometric Progressions

$T_{n} = a r^{n - 1}$

a = first term

r = common ratio

n = the number of term

T_n = the nth term

Example 1:

Find the given term for each of the following geometric progressions.

(a) 8 ,4 ,2 ,...... T₈

(b) $\frac{16}{27}, \frac{8}{9}, \frac{4}{3}, .....$ , T₆

Solution:

$\begin{array}{l} T_{n} = a r^{n - 1} \\ T_{1} = a r^{1 - 1} = a r^{0} = a \leftarrow (First term) \\ T_{2} = a r^{2 - 1} = a r^{1} = a r \leftarrow (S e c o n d term) \\ T_{3} = a r^{3 - 1} = a r^{2} \leftarrow (T h i r d term) \\ T_{4} = a r^{4 - 1} = a r^{3} \leftarrow (Fourth term) \end{array}$

(a)

$\begin{array}{l} 8, 4, 2, ..... \\ a = 8, r = \frac{4}{8} = \frac{1}{2} \\ T_{8} = a r^{7} \\ T_{8} = 8 {(\frac{1}{2})}^{7} = \frac{1}{16} \end{array}$

(b)

$\begin{array}{l} \frac{16}{27}, \frac{8}{9}, \frac{4}{3}, ..... \\ a = \frac{16}{27} \\ r = \frac{T_{2}}{T_{1}} = \frac{\frac{16}{27}}{\frac{8}{9}} = \frac{2}{3} \\ T_{6} = a r^{5} \\ = \frac{16}{27} {(\frac{2}{3})}^{5} = \frac{512}{6561} \end{array}$