Balance of 1000?

Question

How much must you deposit in an account that pays 6% interest compounded yearly to have a balance of $1000 after 8 years?

Can someone tell me the equation for this?

Answer 1

Suppose the sum is 1
after one year you have 1.06
after eight yeras you have (1.06)^8 =1.59
So 1.59D =1000$

D= 627 $

Answer 2

Let P = Principal (initial amount you deposit)
r = annual rate of interest (percentage)
n = number of years the amount is deposited
A = the amount of money accumulated after n years, including interest

The formula for the interest compounded in one year is
A= P(1 + r) n

After 5 years, it would be A=P(1+r)^5

Going to our problem.
A = $ 1000
r = 6% (or 0.06)
n = 8
P =?

A = P(1+r)^n
1000 = P(1+ 0.06) ^8
1000 = P (1.06)^8 ....... 1.06^8 = 1.5938
1000 = P (1.5938)
Divide both sides by 1.5938
1000/1.5938 = P
627.43 = P

Deposit about $ 627.50 to get $ 1000 after 8 years at 6% interest compounded annually

Answer 3

A * (1.06)^8 = 1000
A * 1.593848075 = 1000
A = 1000 / 1.593848075
A = 627.41

Plus or minus up to 4 cents, depending on how the bank rounds off the interest each year.

Answer 4

suppose the amount is x

after one year it is x(1+6/100)

after 2 years x(1+6/100)^2
after 8 years x(1+6/100)^8
this shall be 1000
so eqution is

x(1+6/100)^8 = 1000

Answer 5

amount= original amount x (1+ decimal interest)^number of years
so...
if you put the numbers in and change the subject you get:
1000=y x 1.06 ^8
y=1000/1.06 ^8
and then you just put it into a calculator