Hi Fre,
To save some typing let $A be the amount of the loan, $P the payment you make each year and i be the annual interest rate. At the end of the first year you owe $A(1 + i) and you pay $P so you left owing
$A(1 + i)  $P
Thus at the end of the second year you owe this amount times 1 + i and you pay $P so you are left owing
[$A(1 + i)  $P](1 + i)  $P = $A(1 + i)^{2}  $P[(1 + i ) + 1]
This is the amount you owe at the start of the third year so, at the end of the third year you owe this amount times 1 + i and you make a payment of $P. Likewise in the fourth and fifth years. Thus at the end of the fifth year you owe
$A(1 + i)^{5}  $P[(1 + i)^{4} + (1 + i)^{3} + (1 + i)^{2} + (1 + i ) + 1]
But, at the end of the fifth year you have paid off the loan so this is zero. Hence
$50,000(1.1)^{5}  $P[(1.1)^{4} + (1.1)^{3} + (1.1)^{2} + (1.1 ) + 1] = 0 *
Finally
(1.1)^{4} + (1.1)^{3} + (1.1)^{2} + (1.1 ) + 1
is the sum of the first five terms of a geometric series so use you knowledge of geometric series to evaluate this sum and then solve equation * for P.
Penny
