Question #36bf7

3 Answers
Mar 12, 2016

Answer:

#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.

Mar 12, 2016

Answer:

#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#.

Mar 12, 2016

Answer:

#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#.