Question

Question #36bf7

Answer 1

`n^("th")` `"term"=t_n=2(3)^(n-1)`
number of terms`=7`

Explanation:

Recall that a geometric sequence can be written in the following form:

`color(blue)(|bar(ul(color(white)(a/a)t_n=ar^(n-1)color(white)(a/a)|)))`

where:
`t_n=`term number
`a=`first term
`r=`common ratio
`n=`number of terms

Question 1 i)
`1`. To determine the `n^("th")` term, we must form an equation for the geometric sequence. Start by finding the value of `r`, the common ratio. To do this, divide any term by the term before it. In this case, we will divide the second term by the first term.

`r=t_2/t_1`

`r=6/2`

`r=3`

`2`. Substitute the values of `a` and `r` into the geometric sequence formula.

`t_n=ar^(n-1)`

`color(green)(|bar(ul(color(white)(a/a)t_n=2(3)^(n-1)color(white)(a/a)|)))`

`:.`, the `n^("th")` term is `t_n=2(3)^(n-1)`.

Question 1 ii)
`1`. To determine the number of terms in the sequence, take the formula for the general term derived in part i) and substitute 1458, the last term, as `t_n`.

`t_n=2(3)^(n-1)`

`1458=2(3)^(n-1)`

`2`. Solve for `n`.

`729=3^(n-1)`

`3^6=3^(n-1)`

`6=n-1`

`color(green)(|bar(ul(color(white)(a/a)n=7color(white)(a/a)|)))`

`:.`, there are `7` terms in the sequence.

Answer 2

`a=4`
`r=2`

Explanation:

Question 2
`1`. Using the general formula for a geometric sequence, `t_n=ar^(n-1)`, substitute the second term, `8`, and the fifth term, `64`, along with their other appropriate known values into two separate equations.

`t_n=ar^(n-1)color(white)(XXXXXXXXx)t_n=ar^(n-1)`

`8=ar^(2-1)color(white)(XXXXXXXXX)64=ar^(5-1)`

`8=arcolor(white)(XXXXXXXXXXX)64=ar^4`

`2`. Using elimination, divide `64=ar^4` by `8=ar` to find the value of `r`.

`(64=ar^4)/(8=ar)`

`(64=color(red)cancelcolor(black)ar^4)/(8=color(red)cancelcolor(black)ar)`

`8=r^3`

`r=root(3)(8)`

`color(green)(|bar(ul(color(white)(a/a)r=2color(white)(a/a)|)))`

`3`. Knowing that `r=2`, substitute the value into `8=ar` to find the value of `a`.

`8=ar`

`8=a(2)`

`color(green)(|bar(ul(color(white)(a/a)a=4color(white)(a/a)|)))`

`:.`, the first term is `4` and the common ratio is `2`.

Answer 3

`a=6`
`r=1/3`

Explanation:

Question 3
`1`. Since the first term is represented by `color(orange)a`, the second term by `color(brown)(ar)`, the third term by `color(purple)(ar^2)`, and the fourth term by `color(blue)(ar^3)`, use these terms to create two algebraic expressions.

Equation `1`: `color(orange)a+color(purple)(ar^2)=20/3`

Equation `2`: `color(brown)(ar)+color(blue)(ar^3)=20/9`

`2`. For each equation, solve for `a`.

Equation `1`:`color(white)(XXXXXXXXX)`Equation `2`:

`color(orange)a+color(purple)(ar^2)=20/3color(white)(XXXXXXX)color(brown)(ar)+color(blue)(ar^3)=20/9`

`a(1+r^2)=20/3color(white)(XXXXXX)a(r+r^3)=20/9`

`a=20/(3(1+r^2))color(white)(XXXXXXX)a=20/(9(r+r^3))`

`3`. Since you rearranged the two equations in terms of `a`, then the equations must equal each other. Thus, set the two equations equal to each other and solve for `r`.

`20/(color(turquoise)3(1+r^2))=20/(color(teal)9(r+r^3))`

`(color(turquoise)3*color(teal)9)[20/(color(turquoise)3(1+r^2))]=(color(turquoise)3*color(teal)9)[20/(color(teal)9(r+r^3))]`

`(color(red)cancelcolor(turquoise)3*color(teal)9)[20/(color(red)cancelcolor(turquoise)3(1+r^2))]=(color(turquoise)3*color(red)cancelcolor(teal)9)[20/(color(red)cancelcolor(teal)9(r+r^3))]`

`(9(20))/(1+r^2)=(3(20))/(r+r^3)`

`180/color(magenta)(1+r^2)=(60)/(rcolor(magenta)((1+r^2)))`

`color(magenta)((1+r^2))(180/color(magenta)(1+r^2))=color(magenta)((1+r^2))[(60)/(rcolor(magenta)((1+r^2)))]`

`color(red)cancelcolor(magenta)((1+r^2))(180/color(red)cancelcolor(magenta)(1+r^2))=color(red)cancelcolor(magenta)((1+r^2))[(60)/(rcolor(red)cancelcolor(magenta)((1+r^2)))]`

`180=60/r`

`180r=60`

`color(green)(|bar(ul(color(white)(a/a)r=1/3color(white)(a/a)|)))`

`4`. Knowing that `r=1/3`, substitute this value into either equation `1` or `2` to solve for `a`. In this case, we will use equation `1`.

`a=20/(3(1+r^2))`

`a=20/(3(1+(1/3)^2)`

`a=20/(3(1+1/9))`

`a=20/(3((9+1)/9))`

`a=20/(color(red)cancelcolor(black)3^1((10)/color(red)cancelcolor(black)9^3))`

`a=20/(10/3)`

`a=20*3/10`

`a=color(red)cancelcolor(black)20^2*3/color(red)cancelcolor(black)10^1`

`color(green)(|bar(ul(color(white)(a/a)a=6color(white)(a/a)|)))`

`:.`, `a` is `6` and `r` is `1/3`.