## - Geometric Progressions

A sequence $$a_1, a_2, a_3, … , a_n,…$$ is called a geometric sequence, or geometric progression, if there exists a non-zero constant $$r$$, called the common ratio, such that

$$a_1$$
$$a_2=a_1 r$$
$$a_3=a_2 r$$
$$=a_1 rr$$
$$=a_1 r^2$$
Therefore,
$$a_n=a_1 r^{n-1}, n>1$$ or $$\displaystyle {a_n\over a_{n-1}}=r$$

LESSON 1
Prove that the sequence 1, 3, 9, 27, … is a Geometric Progression.
SOLUTION
• We need to show that $$\displaystyle {a_n\over a_{n-1}}$$ is a constant.
• Determine an expression for $$a^n$$. This can be done by inspection (observation).
$$a_1=3^0, a_2=3^1, a_3=3^2, a_4=3^3$$
$$a_n=3^{n-1}$$
• Use expression for $$a_n$$ to determine an expression for $$a_{n-1}$$
$$a_{n-1}=3^{n-1-1}=3^{n-2}$$
• Show that $$\displaystyle {a_n\over a_{n-1}}$$ is a constant.
$$\displaystyle {a_n\over a_{n-1}}={3^{n-1}\over 3^{n-2}}$$
$$=3^{n-1-(n-2)}$$
$$=3$$

LESSON 2
The first and fourth terms of a geometric progression are 6 and 20.25 respectively. Determine the 8th term of the progression.
SOLUTION
$$a_1=6$$
$$a_4=a_1 r^3=20.25$$
• Solve for $$r$$.
$$\displaystyle r^3={20.25\over6}=3.375$$
$$r=\sqrt[3]{3.375}=1.5$$
• Use the values for $$r$$ and $$a_1$$ to evaluate $$a_8$$.
$$a_8=a_1 r^7$$
$$\displaystyle a_8=6\left({3\over2}\right)^7={6561\over64}$$

LESSON 3
The lengths of the sides of a triangle are in geometric progression and the longest side has a length of 36 cm. Given that the perimeter is 76 cm, find the length of the shortest side.
SOLUTION
• Let longest side be $$a_1$$ and shortest side be $$a_3$$.
$$a_1=36$$
• Use the fact that the perimeter is the sum of the three sides.
$$a_1+a_2+a_3=76$$
Replace terms with equivalent expressions involving $$r$$.
$$a_1+a_1 r+a_1 r^2=76$$
$$36+36r+36r^2=76$$
$$36r^2+36r=40$$
$$9r^2+9r-10=0$$
$$(3r+5)(3r-2)=0$$
$$\displaystyle r={2\over3}$$
Since length cannot be negative
$$\displaystyle a_1 r^2=36\left({2\over3}\right)^2=16$$

SUM OF A GEOMETRIC PROGRESSION
The sum of the first $$n$$ terms of a G.P is given by
$S_n={a_1(1-r^n)\over 1-r}, r<1$
or
$S_n={a_1(r^n-1)\over r-1}, r>1$

LESSON 1
The fourth term of a geometric progression is 6 and the seventh term is $$-48$$. Calculate
(i) the common ratio,
(ii) the first term,
(iii) the sum of the first eleven terms.
SOLUTION
$$a_4=a_1 r^3=6$$
$$a_7=a_1 r^6=-48$$
$$\displaystyle {a_7\over a_4}={a_1 r^6\over a_1 r^3}=r^3$$
$$\displaystyle r^3=-{48\over 6}=-8$$
$$r=-2$$

$$a_1 r^3=6$$
$$a_1 (-8)=6$$
$$\displaystyle a_1=-{3\over4}$$

$$\displaystyle S_n={a_1 (1-r^n )\over 1-r}$$
$$\displaystyle S_{11}={-{3\over4} (1-(-2)^11)\over 1-(-2)}$$
$$S_{11}=-512.25$$

LESSON 2
Given that $$\displaystyle S_n={5\over4} \left(1-\left({1\over3}\right)^n\right)$$, find $$a_n$$ and prove that this sequence is a Geometric Progression.
SOLUTION
$$a_n=S_n-S_{n-1}$$
$$\displaystyle ={5\over4} \left(1-\left({1\over3}\right)^n\right)-{5\over4}\left(1-\left({1\over3}\right)^{n-1}\right)$$
$$\displaystyle ={5\over4}-{5\over4} \left({1\over3}\right)^n-{5\over4}+{5\over4} \left({1\over3}\right)^{n-1}$$
$$\displaystyle =-{5\over4} \left({1\over3}\right)^n+{5\over4} \left({1\over3}\right)^{n-1}$$
• Use laws of Indices
$$\displaystyle =-{5\over4} \left({1\over3}\right)^{n-1} \left({1\over3}\right)+{5\over4} \left({1\over3}\right)^{n-1}$$
• Factor out $$\displaystyle {5\over4} \left({1\over3}\right)^{n-1}$$
$$\displaystyle ={5\over4} \left({1\over3}\right)^{n-1} \left(-{1\over3}+1\right)$$
$$\displaystyle ={5\over4} \left({1\over3}\right)^{n-1} \left({2\over3}\right)$$
$$\displaystyle ={5\over6} \left({1\over3}\right)^{n-1}$$

• Determine $$a_{n-1}$$
$$\displaystyle a_{n-1}={5\over6} \left({1\over3}\right)^{n-2}$$
$$\displaystyle {a_n\over a_{n-1}}={{5\over6}\left({1\over3}\right)^{n-1}\over {5\over6}\left({1\over3}\right)^{n-2}}$$
$$\displaystyle ={1\over3}$$

SUM TO INFINITY
What would be the sum of the infinite series
$$\displaystyle 1+{1\over2}+{1\over4}+{1\over8}+{1\over16}+⋯$$
If we think about it we should realise that the sum appears to be 2. Since the sum appears to tend towards a specific number as it goes on indefinitely we refer to this series as a CONVERGENT series. The sum of this series can be given using the formula
$$\displaystyle S_\infty={a_1\over 1-r}, -1<r<1$$
For our series above we have $$a_1=1$$ and $$\displaystyle r={1\over2}$$, therefore
$$\displaystyle S_\infty={1\over 1-{1\over2}}=2$$
Thus, we see that our intuitive answer is indeed correct.

LESSON 1
The first and fourth terms of a geometric progression are 500 and 32 respectively. Find
(i) the values of second and third terms
(ii) the sum to infinity of the progression
SOLUTION
(i) $$a_1=500$$
$$a_4=32$$
$$a_1 r^3=32$$
• Determine the value of the common ratio, $$r$$.
$$500r^3=32$$
$$\displaystyle r^3={32\over 500}={8\over125}$$
$$\displaystyle r={2\over5}$$

$$a_2=a_1 r$$
$$\displaystyle =500\left({2\over5}\right)$$
$$\displaystyle =200$$
$$\displaystyle a_3=a_1 r^2$$
$$=500\left({2\over5}\right)^2$$
$$=80$$

(ii) $$\displaystyle S_\infty={a\over 1-r}$$
$$\displaystyle ={500\over 1-{2\over5}}$$
$$\displaystyle ={2500\over3}$$

LESSON 2
The first term of a geometric progression is a and the common ratio is $$r$$. Given that $$a=12r$$ and that the sum to infinity is 4, calculate the third term.
SOLUTION
Use $$S_\infty=4$$ to determine $$r$$.
$$\displaystyle S_\infty={a_1\over 1-r}$$
$$\displaystyle 4={12r \over 1-r}$$
$$4-4r=12r$$
$$4=16r$$
$$\displaystyle r={1\over4}$$
• Determine the value of $$a_1$$
$$a_1=12r$$
$$\displaystyle a_1=12\left({1\over4}\right)$$
$$a_1=3$$
• Calculate the value of the third term.
$$a_3=a_1 r^2$$
$$\displaystyle a_3=3\left({1\over4}\right)^2$$
$$\displaystyle a_3={3\over16}$$

LESSON 3
The first term of a geometric series is 120. The sum to infinity of the series is 480. Given that the sum of the first $$n$$ terms is greater than 300, determine the smallest possible value of $$n$$.
SOLUTION
Determine $$r$$.
$$\displaystyle S_\infty={a\over 1-r}$$
$$\displaystyle 480={120\over 1-r}$$
$$480(1-r)=120$$
$$\displaystyle 1-r={1\over4}$$
$$\displaystyle r={3\over4}$$
• Determine $$n$$ as follows:
$$\displaystyle S_n={a(1-r^n)\over 1-r}$$
$$S_n>300$$
$$\displaystyle {120\left(1-\left({3\over4}\right)^n\right)\over 1-{3\over4}}>300$$
$$\displaystyle {120\left(1-\left({3\over4}\right)^n\right)\over {1\over4}}>300$$
$$\displaystyle 480\left(1-\left({3\over4}\right)^n\right)>300$$
$$\displaystyle 1-\left({3\over4}\right)^n>{5\over8}$$
$$\displaystyle \left({3\over4}\right)^n<{3\over8}$$
• Take natural logarithms of both sides.
$$\displaystyle \ln⁡\left({3\over4}\right)^n <\ln\left({3\over8}\right)$$
$$\displaystyle n \ln\left({3\over4}\right)<\ln\left({3\over8}\right)$$
• Divide both sides by $$\displaystyle \ln\left({3\over4}\right)$$. It is important to note that this value is negative.
$$\displaystyle n>{\ln\left({3\over8}\right)\over \ln\left({3\over4}\right)}$$
$$n>3.4$$
$$n=4$$

LESSON 4
Determine whether the geometric series
$$\displaystyle \sum_{r=1}^\infty\left({1\over2}\right)^r$$
is convergent. If it converges, determine its sum.
SOLUTION
• We need to show that $$-1<r<1$$.
• State the first few terms of the series.
$$\displaystyle \sum_{r=1}^\infty \left({1\over2}\right)^r={1\over2}+{1\over4}+{1\over8}+{1\over16}+⋯$$
• Determine the value of $$r$$.
$$\displaystyle r={{1\over4}\over {1\over2}}={1\over2}$$
Since $$-1<r<1$$, the series converges.
$$\displaystyle S_\infty={{1\over2}\over 1-{1\over2}}=1$$