Introduction to Sequences
Introduction to Arithmetic Progressions
Recurrence relationships for AP's
Terms in Arithmetic Progressions
Graphs and Tables - AP's
Notation for a Series
Arithmetic Series (defined limits)
Arithmetic Series (using graphics calculators)
Applications of Arithmetic Progressions
Introduction to Geometric Progressions
Recurrence relationships for GP's
Finding the Common Ratio
Terms in Geometric Progressions
Graphs and Tables - GP's
Geometric Series (using graphics calculators)
Infinite sum for GP's
Applications of Geometric Progressions
Applications of Geometric Series
Sequences and Saving Money (Investigation) LIVE
First Order Linear Recurrences Introduction
Graphs and Tables - Recurrence Relations
Solutions to Recurrence Relations
Steady state solutions to recurrence relations
Applications of Recurrence Relations
Level 7 - NCEA Level 2
Recall that the sum to $n$n terms of a GP is given by:
$S_n=\frac{a\left(1-r^n\right)}{1-r}$Sn​=a(1−rn)1−r​
Image -Nicolas Reusens/Barcroft from
http://www.telegraph.co.uk
Suppose a rare species of frog can jump up to $2$2 metres in one bound. One such frog with an interest in mathematics sits $4$4 metres from a wall. The curious amphibian decides to jump $2$2 metres toward it in a single bound. After it completes the feat, it jumps again, but this time only a distance of $1$1 metre. Again the frog jumps, but only $\frac{1}{2}$12​ a metre, then jumps again, and again, each time halving the distance it jumps. Will the frog get to the wall?
The total distance travelled toward the wall after the frog jumps $n$n times is given by $S_n=\frac{2\left(1-\left(\frac{1}{2}\right)^n\right)}{1-\frac{1}{2}}$Sn​=2(1−(12​)n)1−12​​ .
So after the tenth jump, the total distance is given by $S_{10}=4\left(1-\left(\frac{1}{2}\right)^{10}\right)=4\left(1-\frac{1}{1024}\right)$S10​=4(1−(12​)10)=4(1−11024​)
If we look carefully at the last expression, we realise that, as the frog continues jumping toward the wall, the quantity inside the square brackets approaches the value of $1$1 but will always be less than $1$1. This is the key observation that needs to be made. Therefore the entire sum must remain less than $4$4, but the frog can get as close to $4$4 as it likes simply by continuing to jump according to the geometric pattern described.
Whenever we have a geometric progression with its common ratio within the interval $-1<1$< span=""><1$<>−1<r<1 then, no matter how many terms we add together, the sum will never exceed some number $L$L called the limiting sum. We sometimes refer to it as the infinite sum of the geometrical progression.
Since for any GP, $S_n=\frac{a\left(1-r^n\right)}{1-r}$Sn​=a(1−rn)1−r​ , if the common ratio is within the interval $-1<1$< span=""><1$<>−1<r<1 then, as more terms are added, the quantity $\left(1-r^n\right)$(1−rn) will become closer and closer to $1$1. This means that the sum will get closer and closer to:
$S_{\infty}=\frac{a}{1-r}$S∞​=a1−r​
Checking our frogs progress, $S_{\infty}=\frac{2}{1-\left(\frac{1}{2}\right)}=4$S∞​=21−(12​)​=4 is the limiting sum.
As another example, the limiting sum of the geometric series $108+36+12\dots$108+36+12… is simply $S_{\infty}=\frac{108}{1-\left(\frac{1}{3}\right)}=162$S∞​=1081−(13​)​=162.
Consider the infinite geometric sequence: $2$2, $\frac{1}{2}$12​, $\frac{1}{8}$18​, $\frac{1}{32}$132​, $\ldots$…
Determine the common ratio, $r$r, between consecutive terms.
Find the limiting sum of the geometric series.
Reveal Solution
Consider the infinite geometric sequence: $125$125, $25$25, $5$5, $1$1, $\ldots$…
Consider the infinite geometric sequence: $16$16, $-8$−8, $4$4, $-2$−2, $\ldots$…
Determine the common ratio between consecutive terms.
M7-3
Use arithmetic and geometric sequences and series
Apply sequences and series in solving problems