Answer
a). $A_{n}=3.2000\times \left( 1.05\right) ^{n-1}\approx 30476.19 \times 1.05 ^{n} $
b)
$A_{1}=32.000\$ $
$A_{2}=33600\$ $
$A_{3}=35280\$ $
$A_{4}=37041\$ $
$A_{5}=38896.2\$ $
$A_{6}=40481.01\$ $
$A_{7}\approx 42883.06\$ $
$A_{8}\approx 45027.2\$ $
Work Step by Step
If it increases by $5\% $ we must multiply it by $1.05$
($\dfrac {100\% +5\% }{100\% }$)
So it will create a geometric sequence shown belov
$32.000,3.2000\times 1.05,32.000\times \left( 1.05\right) ^{2}\ldots $
So
nth year salary will be:
$A_{n}=A\times r^{n-1}=3.2000\times \left( 1.05\right) ^{n-1}$
So if we calculate we get
$A_{1}=32.000\$ $
$A_{2}=33600\$ $
$A_{3}=35280\$ $
$A_{4}=37041\$ $
$A_{5}=38896.2\$ $
$A_{6}=40481.01\$ $
$A_{7}\approx 42883.06\$ $
$A_{8}\approx 45027.2\$ $
