- 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.
\(\displaystyle S_\infty={{1\over2}\over 1-{1\over2}}=1\)