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 topay the bank $18 000 per month. The interest rate is 4.5% per annum, compounded monthly, and theloan 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"0.00375c Find the amount owing on the loan at the end of the fifth year, and state whether this is more or lessthan 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 thatlog10 1.5484n =log10 1.00375

(e) The amount borrowed is $1.7 million P=$1.7million=$1700000 Rate of interest, I=4.5% Monthly…

Answered: e If it is paid out after n months…

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 =

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.3: Geometric Sequences

Problem 81E

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Question

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

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