Sequences and Series

Hong Kong

Stage 4 - Stage 5

Lesson

Sometimes we are given just enough information about a geometric progression to find its common ratio. Recall that the $n$`n`th term of a GP is given by:

$t_n=ar^{n-1}$`t``n`=`a``r``n`−1

Note that there are two variables in the formula – the first term $a$`a` and the common ratio $r$`r`. It is often the case in mathematics that in order to find the value of these parameters we need to be given two separate pieces of information.

Carefully consider the following example:

Suppose that for a certain geometric progression we are told that its third term is $63$63 and its fifth term is $567$567. Hence we know that:

$63=ar^2$63=`a``r`2

$567=ar^4$567=`a``r`4

Here we have two equations and two unknowns, so we should be able to find the first term and common ratio of the sequence. We do this by division, so that:

$\frac{567}{63}=\frac{ar^4}{ar^2}$56763=`a``r`4`a``r`2

By cancelling factors this equation becomes $9=r^2$9=`r`2 and so $r=\pm3$`r`=±3 . This means there are two possible geometric progressions to consider.

Firstly, if $r=3$`r`=3, then since we have that $63=9a$63=9`a` and so $a=7$`a`=7. So the first five terms of the sequence become $7,21,63,189$7,21,63,189 and $567$567. Secondly, if $r=-3$`r`=−3 then $a$`a` is again $7$7, but every second term changes sign so that the first five terms become $7,-21,63,-189$7,−21,63,−189 and $567$567. Note that both of these sequences have the third and fifth term $63$63 and $567$567 respectively as required by the original information given in the question.

In a geometric progression, $T_4=54$`T`4=54 and $T_6=486$`T`6=486.

Solve for $r$

`r`, the common ratio in the sequence. Write both solutions on the same line separated by a comma.For the case where $r=3$

`r`=3, solve for $a$`a`, the first term in the progression.Consider the sequence in which the first term is positive. Find an expression for $T_n$

`T``n`, the general $n$`n`th term of this sequence.

In a geometric progression, $T_7=\frac{64}{81}$`T`7=6481 and $T_8=\frac{128}{243}$`T`8=128243.

Find the value of $r$

`r`, the common ratio in the sequence.Find the first three terms of the geometric progression:

$\editable{},\editable{},\editable{}$,,, $\ldots$…, $\frac{64}{81}$6481, $\frac{128}{243}$128243

In a geometric progression, $T_4=-192$`T`4=−192 and $T_7=12288$`T`7=12288.

Find the value of $r$

`r`, the common ratio in the sequence.Find $a$

`a`, the first term in the progression.Find an expression for $T_n$

`T``n`, the general $n$`n`th term.