Find the second term of a geometric sequence whose third term is 4/3, and whose sixth term is -32/27.

Guest Nov 18, 2018

#1**+1 **

Let r be the common ratio

Let the first term = a

So.....the third term is ar^2

And the sixth term is ar^2

So.....we have this ratio

ar^2 (4/3)

____ = _______

ar^5 (-32/27)

r^2 (4/3)

___ = ______ and we can invert these fractions and write

r^5 (-32/27)

r^3 = (-32/27) / (4/3)

r^3 = (-32/27) * ( 3/4) = (-8) / (9)

r = (-8/9)^(1/3) = -2 / (9)^(1/3) = - 2 / (3^2)^(1/3) = - 2 / (3)^(2/3)

So....to find the second term, we have

(2nd term) * r = (3rd term) divide both sides by r

(2nd term ) = (3rd term ) / r

(2nd term) = (4/3) / [ -2 / (3)^(2/3) ] = (4/3) * (3)^(2/3) / -2 =

(4 / -2) * ( 3^(2/3) / 3 ) =

-2 / (3)^1/3

CPhill
Nov 18, 2018