1.6.3 Geometric Progressions - GCEALevelMaths9709

In this topic we will learn how to:

recognise geometric progressions
use the formulae for the $nth$ term and for the sum of the first $n$ terms to solve problems involving geometric progressions
use the condition for convergence of a geometric progression, and the formula for the sum to infinity of a convergent geometric progression

A geometric progression is a sequence of numbers in which each differs from the preceding one by a common ratio. To find the $nth$ term under a geometric progression, we use the formula,
$u_{n} = ar^{n - 1}$ Where $u_{n}$ represents the $nth$ term, $a$ represents the first term of the progression, $n$ represents the position of the $nth$ term and $r$ represents the common ratio.

To find the sum of the first $n$ terms under a geometric progression, we use the following formula,
$S_{n} = \frac{a(1 - r^{n})}{1 - r}$ Where $S_{n}$ represents sum of first $n$ terms, $a$ represents the first term of the progression, $r$ is the common ratio and $r \neq 1$ .
$\textbf{\textcolor{gray}{{Sum to Infinity}}}$ If the common ratio lies between $-1$ and $1$ then the geometric progression converges, and it has a sum to infinity.
To calculate the sum to infinity, we use the formula below,
$S_{\infty} = \frac{a}{1 - r}$ Let’s look at some past paper questions.

1. The sum to infinity of a geometric progression is $9$ times the sum of the first four terms. Given that the first term is $12$ find the value of the fifth term. (9709/12/M/J/19 number 10)

Let’s write out, mathematically, the information we have been given,
$S_{\infty} = 9S_{4}$ We have to use the information above to find $u_{5}$ ,
$u_{5} = ar^{4}$ We already have $a$ so we need to find $r^{4}$ .

Use the formulae for $S_{n}$ and $S_{\infty}$ to substitute,
$S_{\infty} = 9S_{4}$ $\frac{\textcolor{#2192ff}{a}}{1 - r} = \frac{9\textcolor{#2192ff}{a}\left(1 - r^{4}\right)}{1 - r}$ $\frac{\textcolor{#2192ff}{12}}{1 - r} = \frac{9(\textcolor{#2192ff}{12})\left(1 - r^{4}\right)}{1 - r}$ $\frac{12}{1 - r} = \frac{108\left(1 - r^{4}\right)}{1 - r}$ Get rid of the denominator,
$12 = 108(1 - r^{4})$ Expand the brackets and simplify,
$12 = 108 - 108r^{4}$ $108r^{4} = 96$ $r^{4} = \frac{96}{108}$ $r^{4} = \frac{8}{9}$ Now that we have $r^{4}$ we can find $u_{5}$ ,
$u_{5} = \textcolor{#2192ff}{a}\textcolor{red}{r^{4}}$ $u_{5} = \textcolor{#2192ff}{12}\textcolor{red}{\left(\frac{8}{9}\right)}$ $u_{5} = \frac{32}{3}$ Therefore, the final answer is,
$u_{5} = \frac{32}{3}$ 2. The first term of a geometric progression is $\sin^{2}{\theta},$ where $0 < {\theta} < \frac{1}{2}\pi$ . The second term of the progression is $\sin^{2}{\theta} \cos^{2}{\theta}$ . Given that the progression is geometric, find the sum to infinity. (9709/13/M/J/20 number 8)

Let’s write out the information we have been given,
$a = \sin^{2}{\theta}\ \ \ \ \ \ u_{2} = \sin^{2}{\theta} \cos^{2}{\theta}$ The formula for sum to infinity is,
$S_{\infty} = \frac{a}{1 - r}$ We already have $a$ , we need to find $r$ ,
$r = \frac{u_{2}}{u_{1}}$ $r = \frac{\textcolor{#2192ff}{\sin^{2}{\theta}} \cos^{2}{\theta}}{\textcolor{#2192ff}{\sin^{2}{\theta}}}$ Cancel out $\textcolor{#2192ff}{\sin^{2}\theta}$ ,
$r = \cos^{2}{\theta}$ Substitute $a$ and $r$ into the formula for sum to infinity,
$S_{\infty} = \frac{\textcolor{#2192ff}{a}}{1 - \textcolor{red}{r}}$ $S_{\infty} = \frac{\textcolor{#2192ff}{\sin^{2}{\theta}}}{1 - \textcolor{red}{\cos^{2}{\theta}}}$ Use the identity $\sin^{2}{\theta} \equiv 1 - \cos^{2}{\theta}$ to simplify,
$S_{\infty} = \frac{\sin^{2}{\theta}}{\textcolor{#2192ff}{1 - \cos^{2}{\theta}}}$ $S_{\infty} = \frac{\sin^{2}{\theta}}{\textcolor{#2192ff}{\sin^{2}{\theta}}}$ $S_{\infty} = 1$ Therefore, the final answer is,
$S_{\infty} = 1$ 3. The first, second and third terms of a geometric progression are $2p + 6$ , $-2p$ and $p + 2$ respectively, where $p$ is positive. Find the sum to infinity of the progression. (9709/12/O/N/20 number 2)

Let’s write out the information we have been given,
$u_{1} = 2p + 6\ \ \ \ \ \ u_{2} = -2p\ \ \ \ \ \ u_{3} = p + 2$ The above can be written as,
$a = 2p + 6\ \ \ \ \ \ ar = -2p\ \ \ \ \ \ ar^{2} = p + 2$ Note: We have just used the formula for the $nth$ term.

To find the sum to infinity, we need to find $a$ and $r$ by first evaluating $p$ . To evaluate $p$ we can make two equations for $r$ ,
$r = \frac{ar}{a}\ \ \ \ \ \ r = \frac{ar^{2}}{ar}$ $r = \frac{-2p}{2p + 6}\ \ \ \ \ \ r = \frac{p + 2}{-2p}$ Eqaute the two equations together,
$\frac{-2p}{2p + 6} = \frac{p + 2}{-2p}$ Solve for $p$ ,
$\frac{-2p}{2p + 6} = \frac{p + 2}{-2p}$ $(-2p)(-2p) = (p + 2)(2p + 6)$ $4p^{2} = 2p^{2} + 10p + 12$ $4p^{2} - 2p^{2} - 10p - 12 = 0$ $2p^{2} - 10p - 12 = 0$ $p^{2} - 5p - 6 = 0$ $(p + 1)(p - 6) = 0$ $p = -1, p = 6$ $p = 6$ Note: We disregard $p = -1$ because the question states that $p$ is positive.

Substitute $p$ to find $r$ ,
$r = \frac{-2\textcolor{#2192ff}{p}}{2\textcolor{#2192ff}{p} + 6}$ $r = \frac{-2(\textcolor{#2192ff}{6})}{2(\textcolor{#2192ff}{6}) + 6}$ $r = -\frac{2}{3}$ Substitute $p$ to find $a$ ,
$a = 2\textcolor{#2192ff}{p} + 6$ $a = 2(\textcolor{#2192ff}{6}) + 6$ $a = 18$ Now that we have $a$ and $r$ , we can find the sum to infinity,
$S_{\infty} = \frac{\textcolor{#2192ff}{a}}{1 - \textcolor{red}{r}}$ $S_{\infty} = \frac{\textcolor{#2192ff}{18}}{1 - \textcolor{red}{\left(\frac{-2}{3}\right)}}$ $S_{\infty} = 10.8$ Therefore, the final answer is,
$S_{\infty} = 10.8$

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