How do you find the 7th term of the geometric sequence with the given terms #a_2 = 12, a_5 = -768#?

How do you find the 7th term of the geometric sequence with the given terms #a_2 = 12, a_5 = -768#?

2 Answers
Nov 23, 2015

Answer:

First find the common ratio #r#...

Explanation:

#r=[(-768)/12]^(1/(5-2))=-4#

#a_7=a_5xx-4xx-4=-768xx16=-12288#

hope that helped

Nov 23, 2015

Answer:

#a_7= -12288#

Explanation:

Given :
#a_2=12#
#a_5=-768#

Known:
Let a constant be k
Let the geometric ratio be r

#a_2 -> kr^2 = 12..............................(1)# but this in not quite correct!

#a_5# is negative so #kr^5!=+768#

However: if we had #a_n=(-1)^nkr^n# then

#a_5-> (-1)^5kr^5=-768# would be correct

Let:
#a_2=(-1)^2kr^2=12..............................(1_a)#
#a_5=(-1)^5kr^5=-768........................(2)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To find r")#

#color(brown)(Consider: color(white)(...)(Equation (2))/( Equation (1_a))#

#((-1)^5kr^5)/(=(-1)^2kr^2)=(-768)/(12)#

#=>- r^3 =- 64#

Multiply both sides by (-1)

# r^3 = 64#

#color(green)(r= 4 ....................................................(4))#

So #color(green)(a_n=(-1)^nk4^n...................................................(5))#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To find k")#

#color(brown)("Substitute (4) into " (1_a)" giving")#

#a_2=(-1)^2k(4)^2=12#

#color(green)(k=12/16 = 3/4)........................................(6)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To find " a_7)#

Substitute (4) and (6) into (5) giving

#a_n=(-1)^nk4^n -> color(blue)(a_7 = (-1)^7(3/4)(4)^7)#

#a_7 = (-1)(3/4)(16384)#

#color(red)(a_7= -12288)#