Consider the following two cash flow series of payments: Series A is a geometric series increasing at a rate of 9.5% per year. The initial cash payment at the end of year 1 is $1,000. The payments occur annually for 5 years. Series B is a uniform series with payments of value X occurring annually at the end of years 1 through 5. You must make the payments in either Series A or Series B.
a. Determine the value of X for which these two series are equivalent if your TVOM is i = 9%. $
b. If your TVOM is 8%, would you be indifferent between these two series of payments? Enter the PW for each series to support this choice.
c. If your TVOM is 5%, would you be indifferent between these two series of payments? Enter the PW for each series to support this choice.

The present value of series A:[tex]=\Big[1000 \times \dfrac{(1.095)^0}{(1.09)^1}\Big]+\Big[1000 \times \dfrac{(1.095)^1}{(1.09)^2}\Big]+...+\Big[1000 \times \dfrac{(1.095)^4}{(1.09)^5}\Big][/tex]

[tex]= 1000 \Big [ \dfrac{1}{1.09}+ \dfrac{1.095}{1.1881}+ \dfrac{1.199}{1.295}+\dfrac{1.313}{1.912}+\dfrac{1.438}{1.539}\Big][/tex]

[tex]= 1000 \Big[ 0.917 + 0.922 + 0.926 + 0.930 + 0.934\Big][/tex]

[tex]= 1000 \times 4.629[/tex]

[tex]= \$4629[/tex]

Thus, the present value of series A is = $4629

Present value of series A = Present value of series B

[tex]The \ value \ of\ X = \dfrac{Present \ value \ of \ series \ B }{\Big [\dfrac{1-(1+r)^{-n}}{r} \Big ]}[/tex]

[tex]The \ value \ of\ X = \dfrac{4629 }{\Big [\dfrac{1-(1+0.09)^{-5}}{0.09} \Big ]}[/tex]

[tex]The \ value \ of\ X =\dfrac{4629 \times 0.09}{1-0.6499}[/tex]

[tex]The \ value \ of\ X =\dfrac{416.61}{0.3501}[/tex]

[tex]The \ value \ of\ X =1189.97[/tex]

Thus, the value of X = $1189.97

2.

The present value of series A:

[tex]=1000 \times \Big[\dfrac{(1.095)^0}{(1.08)^1}+ \dfrac{(1.095)^1}{(1.08)^2}+...+\dfrac{(1.095)^4}{(1.08)^5}\Big][/tex]

[tex]=1000 \Big [ \dfrac{1}{1.08}+ \dfrac{1.095}{1.1664}+\dfrac{1.199}{1.2597}+\dfrac{1.313}{1.3605}+\dfrac{1.438}{1.4693}\Big ][/tex]

[tex]= 1000\Big [ 0.9259 + 0.9839+0.952 + 0.965+0.979\Big ][/tex]

[tex]= 1000 \times 4.76059[/tex]

[tex]\simeq \$4761[/tex]

Thus, the present value of series A is = $4761

Present value of series B =[tex]Value \ of \ X \times \Big [ \dfrac{1 - (1+r)^{-n} }{r}\Big ][/tex]

[tex]= 1189.97 \times \Big [ \dfrac{1 - (1+0.08)^{-5} }{0.08}\Big ][/tex]

[tex]= \dfrac{1189.97}{0.08} \times \Big ( 1 -0.6806\Big )[/tex]

[tex]= 14874.625 \times 0.3194[/tex]

[tex]= \$4750[/tex]

Thus, the present value of series B = $4750

3.

The present value of series A:

[tex]=1000 \times \Big[\dfrac{(1.095)^0}{(1.05)^1}+ \dfrac{(1.095)^1}{(1.05)^2}+...+\dfrac{(1.095)^4}{(1.05)^5}\Big][/tex]

[tex]=1000 \Big [ \dfrac{1}{1.05}+ \dfrac{1.095}{1.1025}+\dfrac{1.199}{1.1576}+\dfrac{1.313}{1.2155}+\dfrac{1.438}{1.276}\Big ][/tex]

[tex]= 1000\Big [ 0.9524 + 0.9932+1.0357 + 1.08+1.127\Big ][/tex]

[tex]= 1000 \times 5.1883[/tex]

[tex]\simeq \$5,188[/tex]

Thus, the present value of series A = $5188

Present value of series B: =[tex]Value \ of \ X \times \Big [ \dfrac{1 - (1+r)^{-n} }{r}\Big ][/tex]

[tex]= 1189.97 \times \Big [ \dfrac{1 - (1+0.05)^{-5} }{0.05}\Big ][/tex]

[tex]= \dfrac{1189.97}{0.05} \times( 0.2165)[/tex]

[tex]= \$5152.57[/tex]

Thus, the present value of series B = $5153