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Stage 4 - Stage 5

Graphs and Tables - GP's


Recall that every geometric sequence begins as $ar,ar^2,ar^3,...$ar,ar2,ar3,...  and that the $n$nth term is given by:


Having a formula for the $n$nth term allows us to quickly generate a table of values for the sequence. For example in the sequence $12,18,27,\dots$12,18,27, the first term is $12$12 and the common ratio is $1.5$1.5 and so the general term is given by the formula $t_n=12\times\left(1.5\right)^{n-1}$tn=12×(1.5)n1 . By substituting for $n$n appropriately and using a scientific calculator, we can quickly generate the following table of the first $7$7 terms of the sequence:

n 1 2 3 4 5 6
tn 12 18 27 40.5 60.75 91.125

Perhaps more interestingly though is the different types of graphs that geometric sequences correspond to. Usually the graphs are not linear like arithmetic progressions. Graphs of geometric sequences are best known as rising or reducing graphs where the rate of rising continually changes, resulting in a curved growth or decay path. This happens whenever the common ratio is positive like the geometric progression depicted in the above table. However, when the common ratio is negative, the values of successive terms flip their sign so that the graph is depicted as either a growing or diminishing zig-zag path. Think, for example, about the geometric progression that that is identical to the one in the table, but has a negative ratio $r=-1.5$r=1.5 so that its $n$nth term is given by $t_n=12\times\left(-1.5\right)^{n-1}$tn=12×(1.5)n1 . The new table becomes:

n 1 2 3 4 5 6
tn 12 -18 27 -40.5 60.75 -91.125

Checking, for $n=1$n=1, we have $t_1=12\times\left(-1.5\right)^{1-1}=12$t1=12×(1.5)11=12  and for $n=2$n=2 we have $t_2=12\times\left(-1.5\right)^{2-1}=-18$t2=12×(1.5)21=18 so even numbered terms become negative and odd numbered terms become positive. 

Here is a graph of the two geometric sequences depicted in both tables. Note that the odd terms of the zig-zag graph coincide with the terms of the first geometric progression. 

Note also that had the absolute value of the ratios of both geometric progressions been less than 1, then the the absolute value of the terms in both sequences would be reducing in size. 

Worked Examples

Question 1

The $n$nth term of a geometric progression is given by the equation $T_n=2\times3^{n-1}$Tn=2×3n1.

  1. Complete the table of values:

    $n$n $1$1 $2$2 $3$3 $4$4 $10$10
    $T_n$Tn $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
  2. What is the common ratio between consecutive terms?

  3. Plot the points in the table that correspond to $n=1$n=1, $n=2$n=2, $n=3$n=3 and $n=4$n=4.

    Loading Graph...

  4. If the plots on the graph were joined they would form:

    a straight line


    a curved line




A new car purchased for $\$38200$$38200 depreciates at a rate $r$r each year.

  1. Use the values in the table to calculate the rate of change, $r$r.

    years passed ($n$n) $0$0 $1$1 $2$2
    value of car ($A$A) $38200$38200 $37818$37818 $37439.82$37439.82
  2. Write an expression for $A$A for the value of the car $n$n years after it is purchased.

  3. Assuming the rate of depreciation remains constant, how much can the car be sold for after $6$6 years? Give your answer to the nearest cent.

  4. A new motorbike purchased for the same amount depreciates according to the model $V=38200\left(0.97^n\right)$V=38200(0.97n). Which vehicle depreciates more rapidly?






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