Sequences and Series

Lesson

Just as our work with Arithmetic sequences and series is made easier with the use of a graphics calculator or similar technology, so can our work with Geometric sequences and series. Let's take a look at a couple of examples.

The sum of the first $10$10 terms of a geometric sequence is $3774.7459$3774.7459. The sum to infinity of this same sequence is $4000$4000. Determine the first term, $a$`a`, and the common ratio,$r$`r`, for this sequence, given that $r$`r`is positive.

Think: We will use the simultaneous solving facility of our calculator to help us calculate $a$`a`and $r$`r`. To do this, we first need to set up two equations involving both $a$`a`and $r$`r`.

Do:

$3774.7459=\frac{a\left(1-r^{10}\right)}{1-r}$3774.7459=`a`(1−`r`10)1−`r`

$4000=\frac{a}{1-r}$4000=`a`1−`r`

Now we'll type these equations into the simultaneous solver on our calculator.

You'll notice if you scroll across the screen that you'll see two possible pairs of values, but since $r$`r`must be positive we'll only accept the first pair. As the sum of our first $10$10 terms is a rounded value, we'll also round $a$`a`sensibly. Thus $a=1000$`a`=1000 and $r=0.75$`r`=0.75.

The first three terms of a geometric sequence are $x+13$`x`+13, $x+4$`x`+4 and $x-2$`x`−2. Determine the value of the common ratio for this sequence.

Think: Once again we'll make the most of the solving functionality of our calculator to determine the value(s) for $a$`a` which makes these three terms the terms of a geometric sequence. To do this we need to establish the relationship which exists between these three terms.

Do:

Since there must be a common ratio we can say:

$\frac{x+4}{x+13}=\frac{x-2}{x+4}$`x`+4`x`+13=`x`−2`x`+4

Now we'll type this equation in solve to see what $x$`x`value(s) we get.

We can see that there is only one value for $x$`x`that is a solution to this equation.

Now the question asked us not for the value of $x$`x`, but for the value of the common ratio. So we now need to substitute that back into either the left or right side of our equation and then evaluate to find $r$`r`.

$r=\frac{14+4}{14+13}=\frac{18}{27}=\frac{2}{3}$`r`=14+414+13=1827=23

The $6$6th term of a geometric sequence is $557$557 and the $13$13th term is $255642$255642.

Write an equation involving $a$

`a`, the first term, and $r$`r`, the common ratio, of this geometric sequence for the $6$6th term.Write your answer such that the constant term is by itself on one side of the equation.

Write an equation involving $a$

`a`, the first term, and $r$`r`, the common ratio, of this geometric sequence for the $13$13th term.Write your answer such that the constant term is by itself on one side of the equation.

Use the simultaneous solving facility of your calculator to determine the values of $a$

`a`and $r$`r`. Give the value of $a$`a`to the nearest integer and give the value of $r$`r`to one decimal place. Assume $a$`a`and $r$`r`are positive.$a$ `a`$=$= $\editable{}$ $r$ `r`$=$= $\editable{}$ Hence determine the $8$8th term of the sequence.

Give your answer to the nearest integer.

The $4$4th term of a geometric sequence is $384$384 and the sum to infinity of the sequence is $10000$10000.

Write an equation involving $a$

`a`, the first term, and $r$`r`, the common ratio, of this geometric sequence for the $4$4th term.Write your answer such that the constant term is by itself on one side of the equation.

Write an equation involving $a$

`a`, the first term, and $r$`r`, the common ratio, and the sum to infinity of this geometric sequence.Write your answer such that the sum to infinity is by itself on one side of the equation.

One possible solution to these two equations is approximately $a=439$

`a`=439 and $r=0.956$`r`=0.956.Use the simultaneous solving facility of your calculator to determine the exact values of the other solution. Assume $a$

`a`and $r$`r`are positive.$a$ `a`$=$= $\editable{}$ $r$ `r`$=$= $\editable{}$ Hence state the general term, $T_n$

`T``n`, of the sequence.

Use arithmetic and geometric sequences and series

Apply sequences and series in solving problems