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C2 geometric Series Old exam Question

A man borrows a sum of money from a building society and agrees to pay the loan (plus interest) over a period of years. If £A is the sum borrowed and r% the yearly rate of interest charged it can be proved that the amount (£pn) of each annual installment which will extinguish loan in n years is given by the formula

Pn=A(R1)RnRn1 P_{n}= \frac{A(R-1)R^{n}}{R^{n}-1}

where;
R=1+r100 R=1+\frac{r}{100}

a) Assuming this formula calculate Pn P_{n} (correct to the nearest £) if A= 1000, n=25, r=6.5
b) Find in its simplest for this ratio P2n:Pn P_{2n}: P_{n} . Show that this ratio is always greater than 12 \frac{1}{2} and if r=6.5, find the least integral number of years for which this ratio is greater than 35 \frac{3}{5} .

I need help with this question as I dont have a clue on what to do.
I have tried and ended up at a dead end

Thank you for anyone willing to help
(edited 8 years ago)
Original post by bigmansouf

I need help with this question as I dont have a clue on what to do.
I have tried and ended up at a dead end


How far have you got and where are you stuck?
Reply 2
Original post by ghostwalker
How far have you got and where are you stuck?


This what I have done for the long awaited reply i have been at work.
for a
R=1.065 R=1.065
P25=A(R1)RnRn1=1000(1.0651)1.065251.065251=81.98148108 P_{25}=\frac{A(R-1)R^{n}}{R^{n}-1}=\frac{1000(1.065-1)1.065^{25}}{1.065^{25}-1}=81.98148108


for b
P2n:Pn P_{2n}:P_{n}
if n=25 n=25
P50=A(R1)RnRn1=1000(1.0651)1.065501.065501=67.91392551 P_{50}=\frac{A(R-1)R^{n}}{R^{n}-1}=\frac{1000(1.065-1)1.065^{50}}{1.065^{50}-1}=67.91392551

[br]P50:P25[br]=67.91392551:81.98148108[br]=0.8284056917:1 [br]\therefore P_{50}:P_{25}[br]= 67.91392551:81.98148108[br]= 0.8284056917:1
thus the ration shows that it is always greater than 0.5

for the part about the number of years

P2n:Pn>35 P_{2n}:P_{n}> \frac{3}{5}
P2nPn>35 \frac{P_{2n}}{P_{n}}>\frac{3}{5}

A(R1)R2nR2n1A(R1)RnRn1>35 \frac{\frac{A(R-1)R^{2n}}{R^{2n}-1} }{\frac{A(R-1)R^{n}}{R^{n}-1}}>\frac{3}{5}

R2nRn×Rn1R2n1>35 \therefore \frac{R^{2n}}{R^{n}}\times \frac{R^{n}-1}{R^{2n}-1}>\frac{3}{5}

=Rn×Rn1R2n1>35 =R^{n}\times \frac{R^{n}-1}{R^{2n}-1}>\frac{3}{5}

I have done what I can please point me in the right direction

thank you very much
Original post by bigmansouf
This what I have done for the long awaited reply i have been at work.
for a
R=1.065 R=1.065
P25=A(R1)RnRn1=1000(1.0651)1.065251.065251=81.98148108 P_{25}=\frac{A(R-1)R^{n}}{R^{n}-1}=\frac{1000(1.065-1)1.065^{25}}{1.065^{25}-1}=81.98148108


That's fine.


for b
P2n:Pn P_{2n}:P_{n}
if n=25 n=25
P50=A(R1)RnRn1=1000(1.0651)1.065501.065501=67.91392551 P_{50}=\frac{A(R-1)R^{n}}{R^{n}-1}=\frac{1000(1.065-1)1.065^{50}}{1.065^{50}-1}=67.91392551

[br]P50:P25[br]=67.91392551:81.98148108[br]=0.8284056917:1 [br]\therefore P_{50}:P_{25}[br]= 67.91392551:81.98148108[br]= 0.8284056917:1
thus the ration shows that it is always greater than 0.5

for the part about the number of years

P2n:Pn>35 P_{2n}:P_{n}> \frac{3}{5}
P2nPn>35 \frac{P_{2n}}{P_{n}}>\frac{3}{5}

A(R1)R2nR2n1A(R1)RnRn1>35 \frac{\frac{A(R-1)R^{2n}}{R^{2n}-1} }{\frac{A(R-1)R^{n}}{R^{n}-1}}>\frac{3}{5}

R2nRn×Rn1R2n1>35 \therefore \frac{R^{2n}}{R^{n}}\times \frac{R^{n}-1}{R^{2n}-1}>\frac{3}{5}

=Rn×Rn1R2n1>35 =R^{n}\times \frac{R^{n}-1}{R^{2n}-1}>\frac{3}{5}

I have done what I can please point me in the right direction

thank you very much


For part b), I think you've misunderstood what's being asked.

They initially want the ratio P2n:PnP_{2n}:P_n in it's simplest form without putting in values for n,R,A. And show that this ratio is greater than 1/2

Then with r=6.5, do the bit last bit involving 3/5.

One item to note: In your working you ended up with a fraction with R2n1R^{2n}-1 in the denominator. Notice that this is the difference to two squares, and so can be factorised.
Reply 4
Original post by ghostwalker
That's fine.



For part b), I think you've misunderstood what's being asked.

They initially want the ratio P2n:PnP_{2n}:P_n in it's simplest form without putting in values for n,R,A. And show that this ratio is greater than 1/2

Then with r=6.5, do the bit last bit involving 3/5.

One item to note: In your working you ended up with a fraction with R2n1R^{2n}-1 in the denominator. Notice that this is the difference to two squares, and so can be factorised.

thank you
Reply 5
Original post by ghostwalker
That's fine.



For part b), I think you've misunderstood what's being asked.

They initially want the ratio P2n:PnP_{2n}:P_n in it's simplest form without putting in values for n,R,A. And show that this ratio is greater than 1/2

Then with r=6.5, do the bit last bit involving 3/5.

One item to note: In your working you ended up with a fraction with R2n1R^{2n}-1 in the denominator. Notice that this is the difference to two squares, and so can be factorised.



My attempt at part b)i)


P2n:Pn P_{2n}:P_{n}
A(R1)R2nR2n1:A(R1)RnRn1 \frac{A(R-1)R^{2n}}{R^{2n}-1}:\frac{A(R-1)R^{n}}{R^{n}-1}
R2nR2n1:RnRn1 \therefore \frac{R^{2n}}{R^{2n}-1}:\frac{R^{n}}{R^{n}-1}
R2nRn+1:Rn \frac{R^{2n}}{R^{n}+1}: R^{n}
RnRn+1:1 \frac{R^{n}}{R^{n}+1}: 1

shown as required that the ratio is always greater than 0.5

Am i right?


for bii)
using partbi)
RnRn+1:1 \frac{R^{n}}{R^{n}+1}: 1
1.065n1.065n+1:1>35 \frac{1.065^{n}}{1.065^{n}+1}: 1> \frac{3}{5}
1.065n1.065n+1×1>35×1 \frac{1.065^{n}}{1.065^{n}+1} \times 1> \frac{3}{5} \times 1
1.065n1.065n+1>35 \frac{1.065^{n}}{1.065^{n}+1}> \frac{3}{5}

5(1.065n)>3(1.065n+1) 5(1.065^{n})> 3 (1.065^{n} +1)
5(1.065n)>3(1.065n)+3 5(1.065^{n})> 3 (1.065^{n}) +3

5(1.065n)3(1.065n)>3 5(1.065^{n})-3 (1.065^{n})> 3

2(1.065n)>3 2(1.065^{n})> 3

(1.065n)>32 (1.065^{n})> \frac{3}{2}
after taking logs to base 10 and the usual method as shown in c2 logarithms chapter in the edexcel book

n=7
(edited 8 years ago)
Original post by bigmansouf
My attempt at part b)i)


P2n:Pn P_{2n}:P_{n}
A(R1)R2nR2n1:A(R1)RnRn1 \frac{A(R-1)R^{2n}}{R^{2n}-1}:\frac{A(R-1)R^{n}}{R^{n}-1}
R2nR2n1:RnRn1 \therefore \frac{R^{2n}}{R^{2n}-1}:\frac{R^{n}}{R^{n}-1}
R2nRn+1:Rn \frac{R^{2n}}{R^{n}+1}: R^{n}
RnRn+1:1 \frac{R^{n}}{R^{n}+1}: 1

shown as required that the ratio is always greater than 0.5

Am i right?



OK so far.

But you've not justificed why RnRn+1:1 \frac{R^{n}}{R^{n}+1}: 1 as a fraction is greater than 1/2.


for bii)
using partbi)
RnRn+1:1 \frac{R^{n}}{R^{n}+1}: 1
1.065n1.065n+1:1>35 \frac{1.065^{n}}{1.065^{n}+1}: 1> \frac{3}{5}


Mathematically this is a bit odd, but may be OK - don't like it though.
I'd say the ratio a:b, is a/b as a faction and then...


1.065n1.065n+1×1>35×1 \frac{1.065^{n}}{1.065^{n}+1} \times 1> \frac{3}{5} \times 1
1.065n1.065n+1>35 \frac{1.065^{n}}{1.065^{n}+1}> \frac{3}{5}

5(1.065n)>3(1.065n+1) 5(1.065^{n})> 3 (1.065^{n} +1)
5(1.065n)>3(1.065n)+3 5(1.065^{n})> 3 (1.065^{n}) +3

5(1.065n)3(1.065n)>3 5(1.065^{n})-3 (1.065^{n})> 3

2(1.065n)>3 2(1.065^{n})> 3

(1.065n)>32 (1.065^{n})> \frac{3}{2}
after taking logs to base 10 and the usual method as shown in c2 logarithms chapter in the edexcel book

n=7


Working looks fine. Not checked the final value.
Reply 7
Original post by ghostwalker
OK so far.

But you've not justificed why RnRn+1:1 \frac{R^{n}}{R^{n}+1}: 1 as a fraction is greater than 1/2.



to be honest that is the problem i dont know what to do next?

please if possible i need another hint
Original post by bigmansouf
to be honest that is the problem i dont know what to do next?

please if possible i need another hint


One way is partial fractions to get 1- "some fraction", then considering the values that R^n can take show that the something is < 1/2, hence the whole thing is > 1/2.
Reply 9
Original post by ghostwalker
One way is partial fractions to get 1- "some fraction", then considering the values that R^n can take show that the something is < 1/2, hence the whole thing is > 1/2.

thank u

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