How do you find the 7th term of the geometric sequence with the given terms a_2 = 12, a_5 = -768a2=12,a5=768?

How do you find the 7th term of the geometric sequence with the given terms a_2 = 12, a_5 = -768a2=12,a5=768?

2 Answers
Nov 23, 2015

First find the common ratio rr...

Explanation:

r=[(-768)/12]^(1/(5-2))=-4r=[76812]152=4

a_7=a_5xx-4xx-4=-768xx16=-12288a7=a5×4×4=768×16=12288

hope that helped

Nov 23, 2015

a_7= -12288a7=12288

Explanation:

Given :
a_2=12a2=12
a_5=-768a5=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)