Question

On 1st Jan 2000, a man borrowed Rs. A at a fixed compound interest rate of a%. He pays back Rs. P on 1st Jan 2001, and then again Rs. P on 1st Jan 2002. Let a/100 = r

Show that the amount still owing in Rs., just after this second payment, is A(1+r)^2 - P(1+r) - P.

Answer 1

i am trying,
after 1yr : in 2001
he needs to give A(1+r)
pays back P
so, A(1+r) -P
now , you multiply
[ A(1+r) -P] with 1+r for 2002
to get
A(1+r)^2 -P(1+r)
and again he pays back P, so final amount = A(1+r)^2 -P(1+r) -P
phew...

Answer 2

omg

Answer 3

ok wait sub questions coming up bUT THANK YOU

Answer 4

assuming tha the loan is completely paid back after N years, show that

\[P = (Ar)/[1 - (1+r)^-N]\]

Answer 5

ohh wait, i just substitute the sum formula, right?? for GP's??

Answer 6

i guess yes, will have to try

Answer 7

okie.

Answer 8

ok, forget that other question (unless you have the answer, of course :P ) and just see this. Using the same P = blah bah walla formula, we, need to express N in terms of A, P, and r.

Answer 9

means here ? \[P = (Ar)/[1 - (1+r)^{-N}]\]
do u need to isolate N here ?

Answer 10

yup.

Answer 11

let me try
(1-Ar)/P = (1+r)^(-N)

Answer 12

logarithms use karna padega

Answer 13

arrey no yaar. no other method??

Answer 14

and btw no, it would be 1 - (Ar/P) = (1+r)^(-N)

Answer 15

ha wohi :P

Answer 16

without lod, N isolate nai ho sakta.

Answer 17

*log

Answer 18

well, ok.

Answer 19

what u typing so much :P

Answer 20

i think you can prove this by mathematical induction
P=(Ar)/[1−(1+r)^(−N)]
did u try it ?

Answer 21

ummm......nope :P sorry, had gone for dinner.