Remember to register here for FREE to ask any questions you may come across in your QCE studies!The easiest ways to distinguish between arithmetic and geometric sequences is to know what they reflect!

Arithmetic sequences rely on the recursion \(t_{n+1} = t_n + d\). So we
add \(d\) to each term to find the next term in the sequence. If you see a pattern in how we need to constantly add to arrive at subsequent terms, we have an arithmetic sequence.

Geometric sequences rely on the recursion \(t_{n+1} = r t_n\). So we
multiply \(r\) to each term to find the next term in the sequence. Therefore if you see patterns such as doubling, tripling, halving and so on, we have a geometric sequence.
Note of course that \(d\) can be negative, in which case the overall effect is we have subtraction. Similarly \(r\) can be a number between \(1\) and \(1\), in which case the overall effect is division.
Example 1A car is purchased at the end of the year at a net worth of $23000. It is estimated that every year it'll lose 30% of its value. Find by recursion how much its net worth will be 4 years from then.This is a classic example of reducing balance depreciation. Whilst you may or may not have heard the term in your studies, it's expected you recognise that geometric sequences will be useful here! However, you'll want to be a bit careful constructing it.
The depreciation is calculated based off the car's
current worth. But then the depreciation next year will be based off the car's value after one year. Everything is ultimately based off how much the car's value is, not how much has been depreciated! So when we see that 30% of its value is lost, we need to reinterpret it as 70% of its value
remains.
Hence our recurrence relation is \( \boxed{t_{n+1} = 0.7 t_n} \), where \(t_1 = 23000\) (the value at the end of the first year). If we want the value 4 years later, we apply the recursion four times. (Note that four years from now, we're actually in the
fifth year!)
\begin{align*}
t_2 &= 0.7\times23000 = 16100\\
t_3 &= 0.7\times16100 = 11270\\
t_4 &= 0.7\times11270 = 7889\\
t_5 &= 0.7\times 7889 = 5522.3
\end{align*}
So the car will have a value of $5522.30
Example 2 (HSC 2006)
In this problem, we're told that the amount of fruit produced by the orchard decreases by the
same amount every day. That's enough information for us to deduce that we have an
arithmetic sequence here.
The common difference \(d\) will be \(17\). Note that the common difference is computed as \(t_2  t_1 = 543560\), i.e.
next term minus current term! The negative value makes sense, since we produce less fruit each day.
Hence our recurrence is \(t_{n+1} = t_n  17\), with \(t_1 = 560\). Now, trying to find how much is produced on the 14th day will probably take forever. Thankfully, the formula \( \boxed{t_n =t_1 + (n1)d} \) is on your formula sheet, and lets you find the \(n\)th term without excessive recursion!
\[ t_{14} = 560 + 13\times 17 = 339\text{ kg}. \]
We'll skip ii)  they have a nice formula in the HSC to do that part, but we don't in General maths.
Whereas for part iii), we want to check when the production first goes below 60 kg. One way of going about it is to start by finding some value of \(n\), that will set \(t_n\)
equal to 60.
\begin{align*}
t_1 + (n1)d &= 60\\
560 + (n1)(17) &= 60\\
500  17(n1) &= 0\\
500 &= 17(n1)\\
\frac{500}{17} &= n1\\
n &= \frac{500}{17}+1\\
&\approx 30.41
\end{align*}
Then we think intuitively about what goes on. If \(n\) was this number that's almost halfway between \(30\) and \(31\), the harvest would be 60 kg.
Therefore, if \(n=30\), the harvest should still
exceed 60 kg. But if \(n=31\), the harvest will
drop below 60 kg. (Remember  the quantity of fruit the orchard produces is always decreasing!)
Hence, on the 31st day, production first falls below 60kg.
Example 3 (HSC 2005)
This is a nice example that contrasts the two sidebyside. Anne's salary each year forms an arithmetic sequence, whilst Kay's forms a geometric sequence.
The giveaway that Anne's forms an arithmetic sequence is that her earnings always increases by a
fixed amount every year. If we say that we're increasing the salary by $2500, we literally mean that and nothing else. Thus to complete that part of the question, we consider
\[ t_{13} = 50000 + 12\times 2500 = 80000 \]
so Anne's salary will be $80000 in the 13th year.
Whereas the giveaway that Kay's forms a geometric sequence is hidden in the 4%. The question sticks to the wording "in each of the following years her annual salary is increased", but 4% is a proportion taken
relative to the amount in the current year. Because her salary is not the same every year, the amount that it is increased by will no longer be the same every year either, and hence geometric sequences are applicable here.
(In fact, the question is very similar to compound interest.)
Be careful when writing down the common ratio here. Since her salary increases by 4%, the new salary is actually 104% that of the original salary. Hence \(r = 1.04\). Once again, we'll use the shortcut formula instead of recursion  \( t_n = ar^{n1}\) to obtain
\[ t_{13} = 50000\times 1.04^{12} \approx 80051.61 \]
Hence Kay's salary will be $80051.61