e If it is paid out after n months (that is, put A, 0), show that 1.00375 " 1.5484, and hence that log10 1.5484 log10 1.00375 n =

Intermediate Algebra
19th Edition
ISBN:9780998625720
Author:Lynn Marecek
Publisher:Lynn Marecek
Chapter12: Sequences, Series And Binomial Theorem
Section12.3: Geometric Sequences And Series
Problem 12.59TI: New grandparents decide to invest 3200 per month in an annuity for their grandson, The account will...
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Please only show working for part (e), but I thought it may be useful to include the other questions. Thank you!

14 May-Eliane borrowed $1.7 million from the bank to buy some machinery for her farm. She agreed to
pay the bank $18 000 per month. The interest rate is 4.5% per annum, compounded monthly, and the
loan is to be repaid in 10 years.
a Let A, be the balance of the loan after n months. Find a series expression for n.
18000 (1.00375" – 1)
b Hence show that A ,
= 1700000 × 1.00375"
0.00375
c Find the amount owing on the loan at the end of the fifth year, and state whether this is more or less
than half the amount borrowed.
d Find A 120, and hence show that the loan is actually paid out in less than 10 years.
e If it is paid out after n months (that is, put A, = 0), show that 1.00375 "
1.5484, and hence that
log10 1.5484
n =
log10 1.00375
Transcribed Image Text:14 May-Eliane borrowed $1.7 million from the bank to buy some machinery for her farm. She agreed to pay the bank $18 000 per month. The interest rate is 4.5% per annum, compounded monthly, and the loan is to be repaid in 10 years. a Let A, be the balance of the loan after n months. Find a series expression for n. 18000 (1.00375" – 1) b Hence show that A , = 1700000 × 1.00375" 0.00375 c Find the amount owing on the loan at the end of the fifth year, and state whether this is more or less than half the amount borrowed. d Find A 120, and hence show that the loan is actually paid out in less than 10 years. e If it is paid out after n months (that is, put A, = 0), show that 1.00375 " 1.5484, and hence that log10 1.5484 n = log10 1.00375
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