Joe Needs A House

**magentarita** · December 11th 2008, 08:23 PM

Joe wants to buy a house but does not want to get a loan. The average price of the dream house is $500,000 and its price is growing at 5 percent per year. How much should Joe invest in a project at the end of each year for the next 5 years in order to accumulate enough money to buy his dream house with cash at the end of the fifth year? Assume the project pays 12 percent rate of return.

**Soroban** · December 12th 2008, 02:01 PM

Hello, magentarita!

This is a messy one!

Joe wants to buy a house but does not want to get a loan.
The average price of the dream house is $500,000 and its price is growing at 5% per year.
How much should Joe invest in a project at the end of each year for the next 5 years in order
to accumulate enough money to buy his dream house with cash at the end of the 5th year?
Assume the project pays 12% rate of return.

The price of the house is now $500,000.
Its price is increasing by 5% per year.
In five years, its price will be: . $500,\!000(1.05)^5 \:\approx\;\$638,\!140.78$
This is the amount he wants to have at the end of five years.

Let's walk through his investment plan.

At the end of year 1, he invests $x$ dollars.

By the end of year 2, it has grown to $1.12x$ dollars
. . and he deposits another $x$ dollars.
He has $1.12x + x\:=\:(1.12+1)x$ dollars in his account.

By the end of year 3, it has grown to $1.12(1.12 + 1)x \:=\:(1.12^2 + 1.12)x$ dollars
. . and he deposits another $x$ dollars.
He has $(1.12^2 + 1.12)x + x \:=\:(1.12^2 + 1.12 + 1)x$ dollars in his account.

By the end of year 4, it has grown to $1.12(1.12^2+1.12 + 1)x$ dollars
. . and he deposits another $x$ dollars.
He has $(1.12^2 + 1.12 + 1)x + x \:=\:(1.12^3 + 1.12^2 + 1.12 + 1)x$ dollars in his account.

By the end of year 5, it has grown to $1.12(1.12^3+1.12^2+1.12+1)x$ dollars
. . and he deposits another $x$ dollars.
He has $(1.12^4 + 1.12^3 + 1.12^2 + 1.12 + 1)x$ dollars in his account.

And this should equal his desired goal: $638,140.78.

$\text{We have: }\;\underbrace{(1+1.12+1.12^2+1.12^3+1.12^4)}_{\te xt{geometric series}}x \;=\;638,\!140.78$ .[1]

The geometric series has first term $a = 1$ , common ratio $r = 1.12$ , and $n = 5$ terms.
. . Its sum is: . $\frac{1.12^5-1}{1.12 - 1} \:=\:\frac{1.12^5-1}{0.12}$

Then [1] becomes: . $\frac{1.12^5-1}{0.12}\,x \:=\:638,\!140.78 \quad\Rightarrow\quad x \:=\:\frac{0.12}{1.12^5-1}(638,\!140.78)$

Therefore: . $x \;=\;100,\!449.5691 \;\approx\;\boxed{\$100,449.57}$

That is the amount he must invest yearly for 5 years to buy his dream house.

**magentarita** · December 12th 2008, 04:30 PM

Originally Posted by Soroban

Hello, magentarita!

This is a messy one!

The price of the house is now $500,000.
Its price is increasing by 5% per year.
In five years, its price will be: . $500,\!000(1.05)^5 \:\approx\;\$638,\!140.78$
This is the amount he wants to have at the end of five years.

Let's walk through his investment plan.

At the end of year 1, he invests $x$ dollars.

By the end of year 2, it has grown to $1.12x$ dollars
. . and he deposits another $x$ dollars.
He has $1.12x + x\:=\<img src=$ 1.12+1)x" alt="1.12x + x\:=\

1.12+1)x" /> dollars in his account.

By the end of year 3, it has grown to $1.12(1.12 + 1)x \:=\<img src=$ 1.12^2 + 1.12)x" alt="1.12(1.12 + 1)x \:=\

1.12^2 + 1.12)x" /> dollars
. . and he deposits another $x$ dollars.
He has $(1.12^2 + 1.12)x + x \:=\<img src=$ 1.12^2 + 1.12 + 1)x" alt="(1.12^2 + 1.12)x + x \:=\

1.12^2 + 1.12 + 1)x" /> dollars in his account.

By the end of year 4, it has grown to $1.12(1.12^2+1.12 + 1)x$ dollars
. . and he deposits another $x$ dollars.
He has $(1.12^2 + 1.12 + 1)x + x \:=\<img src=$ 1.12^3 + 1.12^2 + 1.12 + 1)x" alt="(1.12^2 + 1.12 + 1)x + x \:=\

1.12^3 + 1.12^2 + 1.12 + 1)x" /> dollars in his account.

By the end of year 5, it has grown to $1.12(1.12^3+1.12^2+1.12+1)x$ dollars
. . and he deposits another $x$ dollars.
He has $(1.12^4 + 1.12^3 + 1.12^2 + 1.12 + 1)x$ dollars in his account.

And this should equal his desired goal: $638,140.78.

$\text{We have: }\;\underbrace{(1+1.12+1.12^2+1.12^3+1.12^4)}_{\te xt{geometric series}}x \;=\;638,\!140.78$ .[1]

The geometric series has first term $a = 1$ , common ratio $r = 1.12$ , and $n = 5$ terms.
. . Its sum is: . $\frac{1.12^5-1}{1.12 - 1} \:=\:\frac{1.12^5-1}{0.12}$

Then [1] becomes: . $\frac{1.12^5-1}{0.12}\,x \:=\:638,\!140.78 \quad\Rightarrow\quad x \:=\:\frac{0.12}{1.12^5-1}(638,\!140.78)$

Therefore: . $x \;=\;100,\!449.5691 \;\approx\;\boxed{\$100,449.57}$

That is the amount he must invest yearly for 5 years to buy his dream house.

Fantastic work as always.

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