r/12 = .06/12 = .005
1+r/12 = 1.005
A = 100 (1.005) [ 1.005^12n -1]/.005
200,000 = 20,000 (1.005)[ 1.005^12n -1]
2000 = [ 1.005^12n -1]
2001 = 1.005^12n
log 2001 = 12 n log 1.005
12 n = 1524.08
n = 127 years
check my arithmetic !
If monthly payments p are deposited in a savings account paying an annual interest rate r, then the amount A in the account after n years is given by
A= p(1+r/12)[(1+r/12)^12n - 1]/ (r/12)
Graph A for the values of p and r.
p = 100, r = 0.06
Estimate n for A = $200,000. (Round your answer to one decimal place.)
1 answer
1 answer