3. The nth Term of Geometric Progressions (Part 2)

Posted on May 24, 2020 by Myhometuition

(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) $\frac{1}{4}, \frac{1}{6}, \frac{1}{9}, ....., \frac{16}{729}$

(c) $- \frac{1}{2}, 1, - 2, ....., 64$

Solution:
(a)

$\begin{array}{l} 2, 4, 8, ....., 8192 \leftarrow (Last term is given) \\ a = 2 \\ r = \frac{T_{2}}{T_{1}} = \frac{4}{2} = 2 \\ T_{n} = 8192 \\ a r^{n - 1} = 8192 \leftarrow (T h e n th term of GP, T_{n} = a r^{n - 1}) \\ (2) {(2)}^{n - 1} = 8192 \\ 2^{n - 1} = 4096 \\ 2^{n - 1} = 2^{12} \\ n - 1 = 12 \\ n = 13 \end{array}$

(b)

$\begin{array}{l} \frac{1}{4}, \frac{1}{6}, \frac{1}{9}, ....., \frac{16}{729} \\ a = \frac{1}{4}, r = \frac{\frac{1}{6}}{\frac{1}{4}} = \frac{2}{3} \\ T_{n} = \frac{16}{729} \\ a r^{n - 1} = \frac{16}{729} \\ (\frac{1}{4}) {(\frac{2}{3})}^{n - 1} = \frac{16}{729} \\ {(\frac{2}{3})}^{n - 1} = \frac{16}{729} \times 4 \\ {(\frac{2}{3})}^{n - 1} = \frac{64}{729} \\ {(\frac{2}{3})}^{n - 1} = {(\frac{2}{3})}^{6} \\ ∴ n - 1 = 6 \\ n = 7 \end{array}$

(c)

$\begin{array}{l} - \frac{1}{2}, 1, - 2, ....., 64 \\ a = - \frac{1}{2}, r = \frac{- 2}{1} = - 2 \\ T_{n} = 64 \\ a r^{n - 1} = 64 \\ (- \frac{1}{2}) {(- 2)}^{n - 1} = 64 \\ {(- 2)}^{n - 1} = 64 \times - 2 \\ {(- 2)}^{n - 1} = - 128 \\ {(- 2)}^{n - 1} = {(- 2)}^{7} \\ n - 1 = 7 \\ n = 8 \end{array}$

(E) Three consecutive terms of a geometric progression

If e, f and g are 3 consecutive terms of GP, then

$\frac{g}{f} = \frac{f}{e}$

Example 3:

If p + 20, p − 4, p −20 are three consecutive terms of a geometric progression, find the value of p.

Solution:

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