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