CAS header

 

Find this content useful?
Think it should stay ad-free?

Geometric Sequences

Geometric Sequences

A sequence is a geometric sequence if it has the form $$ a, ar, ar^2, ar^3, \dots $$ where both a and r are constants. The $n^{th}$ term is given by $$ U_n = ar^{n-1} $$ The sum of the first $n$ terms is given by $$ S_n = \frac{a(1 - r^n)}{1-r} = \frac{a(r^n-1)}{r-1} $$ If $r$, the common ratio, satisfies $|r|\lt 1$, then $$ S_\infty = \frac{a}{1-r} $$

Geometric Sequences 1

On the 1st January each year, £1000 is invested in a high interest account. The interest is 7%, paid annually, on the last day of the year. If the first investment was made on the 1st January 2012, determine the amount of money in the account on 31st December 2021.
solution - press button to display
A useful trick when completing a calculation like this is to draw up a simple table that maps the year, $y$ to the number of term in the sequence $n$ $$ \begin{array}{|c|ccccc|} \hline y & 2012 & 2013 & 2014 & \dots & 2021 \\ \hline n & 1 & 2 & 3 & \dots & 10 \\ \hline \end{array} $$ It is also useful to convince ourselves that we have a geometric sequence. So
Day 1 of 2012, we have $1000$
Day 1 of 2013 we have $1000\times 1.07 + 1000$
Day 1 of 2014 we have $1000\times 1.07^2 + 1000\times 1.07 + 1000$
This is clearly a geometric sequence, with $a=1000$, $r = 1.07$ On day 1 of 2021, we will have $$ S_{10} = \frac{a(r^{10}-1)}{r-1} = \frac{1000(1.07^{10} - 1)}{1.07-1} = 13816.40 $$ On the last day of 2021, we will have $$ 1.07\times S_{10} = 14783.60 $$

Geometric Sequences 2

A sequence begins $k,\;2k+1,3k+14,\dots$ Given that the sequence is a geometric progression, determine the possible values of $k$
solution - press button to display
Since the sequence is a GP, then $\frac{U_2}{U_1} = \frac{U_3}{U_2} = r$ This gives $$ \frac{2k+1}{k} = \frac{3k+14}{2k+1} \Rightarrow k(3k+14) = (2k+1)^2 $$ Solving this quadratic equation gives us $$ k = 5\pm 2\sqrt{6} $$