Stuck with geometric series question: £50,000 is borrowed for a house....
Can't fathom out geometric series question despite knowing the formulas!
The question: £50,000 is borrowed for a house. At end of each year 8% interest is charged on amount still owed. A fixed amount (£P) is payed at the end of each year so that the amount owed after first year is: (50,000 x 1.08) - P
so after third payment it would be (50,000 x 1.08^3) -P(1+1.08+1.08^2). (n less than or equal to 30)
Formula for nth payment would then be: (50,000 X1.08^n) - P(1+1.08+1.08^2.......1.08^n-1). I understand this SO FAR.
The question then asks: ' Use the formula for the sum of a geometric progression to simplify the formula and use this to find the amount £P with n=30 giving answer to nearest penny.
As these are finite divergent series I thin I should use a(r^n-1)/(r-1) but not sure how unless I use this:
[50,000(1.08^30-1)/0.08] - [P(1.08^29-1/0.08)] ??? Not sure what to do here!
Any help-much thanks!
Can't fathom out geometric series question despite knowing the formulas!
The question: £50,000 is borrowed for a house. At end of each year 8% interest is charged on amount still owed. A fixed amount (£P) is payed at the end of each year so that the amount owed after first year is: (50,000 x 1.08) - P
so after third payment it would be (50,000 x 1.08^3) -P(1+1.08+1.08^2). (n less than or equal to 30)
Formula for nth payment would then be: (50,000 X1.08^n) - P(1+1.08+1.08^2.......1.08^n-1). I understand this SO FAR.
The question then asks: ' Use the formula for the sum of a geometric progression to simplify the formula and use this to find the amount £P with n=30 giving answer to nearest penny.
As these are finite divergent series I thin I should use a(r^n-1)/(r-1) but not sure how unless I use this:
[50,000(1.08^30-1)/0.08] - [P(1.08^29-1/0.08)] ??? Not sure what to do here!
Any help-much thanks!