1.3 Geometric sequences and seriesIB Maths: Applications and Interpretation SL: Revision notes

In a geometric sequence each term is found by multiplying the previous term by a fixed number, the common ratio $r$: $u_{n+1}=u_n\times r$, so $r=\frac{u_{n+1}}{u_n}$. The $n$th term is $u_n=u_1r^{n-1}.$ Example: $4,12,36,\ldots$ has $u_1=4$ and $r=3$, so $u_6=4\times3^{5}=972$. Given two terms, divide: $\frac{u_5}{u_2}=r^{3}$. If $u_2=12$ and $u_5=96$ then $r^{3}=8$, so $r=2$ and $u_1=6$.

The sum of the first $n$ terms is $S_n=\frac{u_1\left(r^{n}-1\right)}{r-1}=\frac{u_1\left(1-r^{n}\right)}{1-r},\quad r\ne1.$ Example: for $u_1=4$, $r=3$, $n=6$: $S_6=\frac{4\left(3^{6}-1\right)}{2}=1456$. Use the first form when $r>1$ and the second when $|r|<1$; both give the same value. To find the least $n$ for which $S_n$ passes a target, use a table of values or a graph on your GDC.

$\sum_{k=1}^{n}u_k$ means $u_1+u_2+\cdots+u_n$. For example $\sum_{k=1}^{5}2\times3^{k-1}=2+6+18+54+162=242$, a geometric series with $u_1=2$ and $r=3$. A spreadsheet or GDC can generate terms using 'previous term $\times\,r$' and show sums. If you use technology in an exam you must still identify $u_1$ and $r$ in your working. To find $n$ in $u_1r^{n-1}>T$, compare values in a table or find the intersection of two graphs. Logarithms also work but are not needed for this course.

A percentage increase or decrease each period gives a geometric sequence. For an increase of $p\%$, $r=1+\frac{p}{100}$; for a decrease of $p\%$, $r=1-\frac{p}{100}$. A 4% salary rise has $r=1.04$; an 8% fall in a population has $r=0.92$. If $r>1$ the terms grow, and if $0<r<1$ they decay. Example: a salary of 36 000 AED rising 4% a year is $36\,000\times1.04^{n-1}$ in year $n$; in year 10 it is $36\,000\times1.04^{9}=51\,239.23$ AED.

Geometric sequences model spread of disease, salary increases and decreases, and population growth. Money answers are given to 2 decimal places; other answers to 3 significant figures unless told otherwise. A geometric sequence is an exponential function of $n$, $u_n=u_1r^{n-1}=\frac{u_1}{r}r^{n}$, which links to the exponential models in topic 2. Exponential regression on a GDC (topic 4) gives a model $y=ab^{x}$ where $b$ plays the role of $r$. Compare models: a constant increase (arithmetic) and a constant percentage increase (geometric) can look similar for a few years, but the geometric model eventually grows faster. Over 10 years, 36 000 AED rising 4% a year earns $432\,219.86$ AED, just above a rise of 1600 AED a year, which earns $432\,000$ AED.