Question

$1000 is invested at the beginning of each year for 10 years. The rate of interest is fixed at 7.5% per annum. Interest is compounded annually. Calculate the total value of the investments at the end of the ten years.

Answer 1

 Presumably you are saying that the 7.5% is the actual real effective per annum rate since you did not include a compounding interval. Therefore if you let R = 1.075 which is 7.5%, X = $1,000 and F(n) be the balance after n years then you can figure it out by thinking about what happens. After one year, your balance is: 

F(1) = X * R 

After two years, your balance is: 

F(2) = ( F(1) +X ) * R 
.: 
F(2) = X * R^2 + X * R 

After three years, your balance is: 

F(3) = ( F(2) + X ) * R 
F(3) = X * R^3 + X * R^2 + X * R 

You can see that the general form is: 

F(n) = X * summation of R^k for k = 1 to n 

Applying the summation of a geometric sequence equation and you get: 

F(n) = X * ( ( 1 - R^(n+1) ) / ( 1 - R ) - 1 ) 

Therefore the total value of the investments after 10 years is: 

F(10) = $1,000 * ( ( 1 - 1.075^11 ) / ( 1 - 1.075 ) - 1 ) 
F(10) = $15,208.12 

If you blindly use the programs in the business calculators, you'll get the wrong answer of $14,147.09 because those calculators assume that the deposit is made at the end of the year not at the beginning. The equation is slightly easier to work out that way as the summation of the geometric sequence would be for k from 0 to n-1. That's one of the downfalls of relying on those calculators, you don't actually understand what's happening. 

Now the portion of that which is from the first $1,000 payment is $1,000 * 1.075^10 which is $2,061.03