If a[1]=4 and r=2, what are the first four terms of the geometric sequence?

1 Answer
Mar 13, 2016

Answer:

#4, 8, 16, 32#

Explanation:

Recall that the geometric sequence formula is written as:

#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

Determining the First Four Terms
#1#. Since the value of #a# has already been given, the first term is #4#.

#color(green)(|bar(ul(color(white)(a/a)t_1=4color(white)(a/a)|)))#

#color(red)(rArr)#Sequence thus far: #4,...#

#2#. Using the geometric sequence formula, substitute your known values to determine the second term.

#t_n=ar^(n-1)#

#t_2=4(2)^(2-1)#

#t_2=4(2)^1#

#color(green)(|bar(ul(color(white)(a/a)t_2=8color(white)(a/a)|)))#

#color(red)(rArr)#Sequence thus far: #4,8,...#

#3#. Repeat for the third term.

#t_n=ar^(n-1)#

#t_3=4(2)^(3-1)#

#t_3=4(2)^2#

#t_3=4(4)#

#color(green)(|bar(ul(color(white)(a/a)t_3=16color(white)(a/a)|)))#

#color(red)(rArr)#Sequence thus far: #4,8,16,...#

#4#. Repeat for the fourth term.

#t_n=ar^(n-1)#

#t_4=4(2)^(4-1)#

#t_4=4(2)^3#

#t_4=4(8)#

#color(green)(|bar(ul(color(white)(a/a)t_4=32color(white)(a/a)|)))#

#color(red)(rArr)#Sequence thus far: #4, 8, 16, 32#

#:.#, the first four terms of the sequence are #4, 8, 16, "and"# #32#.