Answer
$a_{5} = \frac{1}{2}$
Work Step by Step
$a_{n}$ of the geometric sequence is $a_{n} = a_{1} . r^{ (n-1)}$
Given $a_{4} =1$
$a_{7} = \frac{1}{8}$
$a_{4} = a_{1} . r^{ (4-1)}$
$a_{4} = a_{1} . r^{ (3)}$
$1 = a_{1} . r^{ 3}$
$a_{1} = \frac{1}{r^{ 3}} $ Equation $(1)$
$a_{7} = a_{1} . r^{ (7-1)}$
$a_{7} = a_{1} . r^{ (6)}$
$\frac{1}{8} = a_{1} . r^{ 6}$
$a_{1} = \frac{1}{8r^{ 6}} $ Equation $(2)$
From Equation $(1)$ and Equation $(2)$
$\frac{1}{r^{ 3}} = \frac{1}{8r^{ 6}} $
Multiply by $r^{ 3}$ both sides.
$\frac{1}{r^{ 3}} \times r^{ 3} = \frac{1}{8r^{ 6}} \times r^{ 3}$
$1 = \frac{1}{8r^{ 3}} $
$r^{ 3} = \frac{1}{8} $
$r = \frac{1}{2} $
Substituting $r$ value in Equation $(1)$
$a_{1} = \frac{1}{(\frac{1}{2})^{ 3}} $
$a_{1} = \frac{1}{(\frac{1}{8})} $
$a_{1} = 8 $
Using $ a_{1} $ , $r$ values and $n=5$ ,
$a_{5} = a_{1} . r^{ (5-1)}$
$a_{5} = 8 (\frac{1}{2} )^{4}$
$a_{5} = 8 (\frac{1}{16} )$
$a_{5} = \frac{1}{2}$
