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Financial Maths, 2ADV M1 2025 HSC 17

A borrower obtains a reducing-balance loan of $800 000 to buy a house.

Interest is charged at 0.5% monthly, compounded monthly.

On the last day of each month, interest is added to the balance owing on the loan and then the monthly repayment of $5740 is made.

Let \(\$ A_n\) be the balance owing on the loan at the end of \(n\) months.

  1. Show that  \(A_2=800\,000(1.005)^2-5740(1.005)-5740\).   (2 marks)

  2. Show that  \(A_n=1\,148\,000-348\,000(1.005)^n\).   (3 marks)

  3. After how many months will the balance owing on the loan first be less than $400 000?   (2 marks)

Show Answers Only

a.   \(\text{See Worked Solutions.}\)

b.   \(\text{See Worked Solutions.}\)

c.   \(\text{154 months}\)

Show Worked Solution
a.     \(A_1\) \(=800\,000(1.005)-5740\)
  \(A_2\) \(=A_1 \times (1.005)-5470\)
    \(=[800\,000(1.005)-5740](1.005)-5740\)
    \(=800\,000(1.005)^2-5740(1.005)-5470\)

 

b.     \(A_3\) \(=800\,000(1.005)^3-5740(1.005)^2-5740(1.005)-5740\)
  \(A_n\) \(=800\,000(1.005)^n-5740(1.005)^{n-1} \ldots -5740(1.005)-5740\)
    \(=800\,000(1.005)^n-5740 \underbrace{\left(1.005+1.005^2+\cdots+1.005^{n-1}\right)}_{\text {GP where} \  a=1 ,  r=1.005}\)
    \(=800\,000(1.005)^n-5740\left(\dfrac{a\left(r^n-1\right)}{r-1}\right)\)
    \(=800\,000(1.005)^n-5740\left(\dfrac{1.005^n-1}{0.005}\right)\)
    \(=800\,000(1.005)^n-1\,148\,000\left(1.005^n-1\right)\)
    \(=1\,148\,000-1\,148\,000(1.005)^n+800\,000(1.005)^n\)
    \(=1\,148\,000-348\,000(1.005)^n\)

 

c.    \(\text{Find} \  n \ \text{such that} \ A_n < 400\,000:\)

\(1\,148\,000-348\,000(1.005)^n\) \(< 400\,000\)
\(348\,000(1.005)^n\) \(>1\,148\,000-400\,000\)
\(1.005^n\) \(>\dfrac{748\,000}{348\,000}\)
\(n \times \ln 1.005\) \(>\ln \left(\frac{187}{87}\right)\)
\(n\) \(>\dfrac{\ln \left(\frac{187}{87}\right)}{\ln 1.005}\)
\(n\) \(>153.42\)

 
\(\therefore \ \text{Balance owing is less than \$400 000 after 154 months.}\)

Financial Maths, 2ADV M1 2022 HSC 32

In a reducing-balance loan, an amount `$P` is borrowed for a period of `n` months at an interest rate of 0.25% per month, compounded monthly. At the end of each month, a repayment of `$M` is made. After the `n`th repayment has been made, the amount owing, `$A_n`, is given by

`A_(n)=P(1.0025)^(n)-M(1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^(n-1))`

(Do NOT prove this.)

  1. Jane borrows $200 000 in a reducing-balance loan as described.

  2. The loan is to be repaid in 180 monthly repayments.

  3. Show that  `M` = 1381.16, when rounded to the nearest cent.  (2 marks)

  4. After 100 repayments of $1381.16 have been made, the interest rate changes to 0.35% per month.

  5. At this stage, the amount owing to the nearest dollar is $100 032. (Do NOT prove this.)

  6. Jane continues to make the same monthly repayments.

  7. For how many more months will Jane need to make full monthly payments of $1381.16?  (3 marks)

  8. The final payment will be less than $1381.16.
  9. How much will Jane need to pay in the final payment in order to pay off the loan?   (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `83`
  3. `$931.54`
Show Worked Solution

a.   `text{Show}\ \ M=$1381.16`

`A_(n)` `=P(1.0025)^(n)-M(1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^(n-1))`  
`0` `=200\ 000(1.0025)^180-M underbrace((1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^179))_(text(GP where)\ a = 1,\ r = 1.0025,\ n = 180)`  
`0` `=200\ 000(1.0025)^180-M((1(1.0025^180- 1))/(1.0025-1))`  

`M((1.0025^180-1)/(1.0025-1))=200\ 000(1.0025)^180`

`:.M` `=200\ 000(1.0025)^180 -: (1.0025^180-1 )/(1.0025-1)`   
  `=1381.163…`  
  `=$1381.16\ \ text{… as required}`  

 

b.   `P=$100\ 032,\ \ r=1.0035  and M=$1381.16`

`text{Find}\ \ n\ \ text{when}\ \ A_n=0:`

`A_(n)` `=P(1.0035)^(n)-1381.16(1+(1.0035)^(1)+(1.0035)^(2)+cdots+(1.0035)^(n-1))`  
`0` `=100\ 032(1.0035)^n-1381.16 ((1.0035^n- 1)/(1.0035-1))`  
`0` `=100\ 032(1.0035)^n-1381.16/0.0035 (1.0035^n- 1)`  
  `=100\ 032(1.0035)^n-394\ 617(1.0035)^n- 394\ 617`  
`294\ 585(1.0035)^n` `=394\ 617`  
`1.0035^n` `=(394\ 617)/(294\ 585)`  
`n` `=ln((394\ 617)/(294\ 585))/ln1.0035`  
  `=83.674…`  
  `=83\ text{more months with full payment}`  

 


♦♦ Mean mark part (b) 38%.

c.   `text{Find}\ \ A_83:`

`A_83` `=100\ 032(1.0035)^83-1381.16 ((1.0035^83- 1)/(1.0035-1))`  
  `=928.291…`  

 
`text{Interest will be added for the last month:}`

`:.\ text{Final payment}` `=928.291… xx 1.0035`  
  `=$931.54`  

♦♦♦ Mean mark part (c) 14%.

Financial Maths, 2ADV M1 2015 HSC 14c

Sam borrows $100 000 to be repaid at a reducible interest rate of 0.6% per month. Let  `$A_n`  be the amount owing at the end of  `n`  months and  `$M`  be the monthly repayment.

  1. Show that  `A_2 = 100\ 000 (1.006)^2 - M (1 + 1.006).`  (1 mark)

  2. Show that  `A_n = 100\ 000 (1.006)^n - M (((1.006)^n - 1)/0.006).`  (2 marks)

  3. Sam makes monthly repayments of $780. Show that after making 120 monthly repayments the amount owing is $68 500 to the nearest $100.  (1 mark)

Immediately after making the 120th repayment, Sam makes a one-off payment, reducing the amount owing to $48 500. The interest rate and monthly repayment remain unchanged.

  1. After how many more months will the amount owing be completely repaid?  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(Show Worked Solutions)}`
  2. `text(Proof)\ \ text{(Show Worked Solutions)}`
  3. `text(Proof)\ \ text{(Show Worked Solutions)}`
  4. `79\ text(months)`
Show Worked Solution
i.  `A_1` `= 100\ 000 (1.006) – M`
`A_2` `= A_1 (1.006) – M`
  `= [100\ 000 (1.006) – M] (1.006) – M`
  `= 100\ 000 (1.006)^2 – M (1.006) – M`
  `= 100\ 000 (1.006)^2 – M (1 + 1.006)\ \ text(…  as required)`

 

ii.  `A_3 = 100\ 000 (1.006)^3 – M (1 + 1.006 + 1.006^2)`

`vdots`

`A_n = 100\ 000 (1.006)^n – M (1 + 1.006 + … + 1.006^(n-1))`

`=> text(S)text(ince)\ \ (1 + 1.006 + … + 1.006^(n-1))\ text(is a)`

`text(GP with)\ \ a = 1,\ r = 1.006`

`:.\ A_n` `= 100\ 000 (1.006)^n – M ((a (r^n – 1))/(r – 1))`
  `= 100\ 000 (1.006)^n – M ((1 (1.006^n – 1))/(1.006 – 1))`
  `= 100\ 000 (1.006)^n – M (((1.006)^n – 1)/0.006)`

`text(…  as required.)`

 

iii.  `text(If)\ \ M = 780 and n = 120`

`A_120` `= 100\ 000 (1.006)^120 – 780 ((1.006^120 – 1)/0.006)`
  `= 205\ 001.80… – 780 (175.0030…)`
  `= 205\ 001.80… – 136\ 502.34…`
  `= 68\ 499.45…`
  `= $68\ 500\ \ text{(to nearest $100)  …  as required}`

 

iv.  `text(After the one-off payment, amount owing)=$48\ 500`

`:. A_n = 48\ 500 (1.006)^n – 780 ((1.006^n – 1)/0.006)`

`text(where)\ \ n\ \ text(is the number of months after)`

 `text(the one-off payment.)`
 

`text(Find)\ \ n\ \ text(when)\ \ A_n = 0`

`48\ 500 (1.006)^n – 780 ((1.006^n – 1)/0.006) = 0`

`48\ 500 (1.006)^n` `= 780 ((1.006^n – 1)/0.006)`
`48\ 500 (1.006)^n` `= 130\ 000 (1.006^n – 1)`
  `= 130\ 000 (1.006)^n – 130\ 000`
`81\ 500 (1.006)^n` `= 130\ 000`
`1.006^n` `= (130\ 000)/(81\ 500)`
`n xx ln 1.006` `= ln\ 1300/815`
`n` `= (ln\ 1300/815)/(ln\ 1.006)`
  `= 78.055…`

 
`:.\ text(The amount owing will be completely repaid after)`

`text(another 79 months.)`

Financial Maths, 2ADV M1 2005 HSC 8c

Weelabarrabak Shire Council borrowed $3 000 000 at the beginning of 2005. The annual interest rate is 12%. Each year, interest is calculated on the balance at the beginning of the year and added to the balance owing. The debt is to be repaid by equal annual repayments of $480 000, with the first repayment being made at the end of 2005.

Let  `A_n`  be the balance owing after the `n`-th repayment.

  1. Show that  `A_2 = (3 × 10^6)(1.12)^2 - (4.8 × 10^5)(1 + 1.12)`.  (1 mark)

  2. Show that  `A_n = 10^6[4 − (1.12)^n]`.  (2 marks)

  3. In which year will Weelabarrabak Shire Council make the final repayment?  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `2017`
Show Worked Solution
i.    `A_1` `= (3 xx 10^6)(1.12) − (4.8 xx 10^5)`
  `A_2` `= A_1(1.12) − (4.8 xx 10^5)`
    `= [(3 xx 10^6)(1.12) − (4.8 xx 10^5)](1.12) − (4.8 xx 10^5)`
    `= (3 xx 10^6)(1.12)^2 − (4.8 xx 10^5)(1.12) − (4.8 xx 10^5)`
    `= (3 xx 10^6)(1.12)^2 − (4.8 xx 10^5)(1 + 1.12)\ \ …text(as required)`

 

ii.  `text(Show)\ A_n = 10^6[4 − (1.12)^n]`

`A_3` `= (3 xx 10^6)(1.12)^3 − (4.8 xx 10^5)(1 + 1.12 + 1.12^2)`
`vdots`  
`A_n` `= (3 xx 10^6)(1.12)^n − (4.8 xx 10^5)(1 + 1.12 + … + 1.12^(n − 1))`
  `=> text(Noting that)\ \ (1+1.12+ … + 1.12^(n-1))\ text(is a)`
  `text(GP where)\ \ a = 1, \ \ r = 1.12`
 `A_n` `= (3 xx 10^6)(1.12)^n − (4.8 xx 10^5)((1(1.12^n − 1))/(1.12 − 1))`
  `= (3 xx 10^6)(1.12)^n − (4.8 xx 10^5)((1(1.12^n − 1))/0.12)`
  `= (3 xx 10^6)(1.12)^n − (4 xx 10^6)(1.12^n − 1)`
  `= 10^6[3(1.12)^n − 4(1.12^n − 1)]`
  `= 10^6[3(1.12)^n − 4(1.12^n) + 4]`
  `= 10^6[4 − (1.12)^n]\ \ …text(as required)`

 

iii.  `text(Find)\ n\ text(such that)\ A_n = 0`

`10^6[4 − (1.12)^n]` `= 0`
`4 − (1.12)^n` `= 0`
`1.12^n` `= 4`
`n xx ln 1.12` `= ln4`
`n` `= (ln4)/ln1.12`
  `= 12.23…\ \ text(years.)`

 

`:.\ text(The final repayment will be made)`

`text(in 2017, the thirteenth year.)`

Financial Maths, 2ADV M1 2006 HSC 8b

Joe borrows $200 000 which is to be repaid in equal monthly instalments. The interest rate is 7.2% per annum reducible, calculated monthly.

It can be shown that the amount, `$A_n`, owing after the `n`th repayment is given by the formula:

`A_n = 200\ 000r^n - M(1 + r + r^2 + … + r^(n-1))`,

where  `r = 1.006`  and  `$M`  is the monthly repayment.  (Do NOT show this.)

  1. The minimum monthly repayment is the amount required to repay the loan in 300 instalments.

     

    Find the minimum monthly repayment.  (3 marks)

  2. Joe decides to make repayments of $2800 each month from the start of the loan.

     

    How many months will it take for Joe to repay the loan?  (2 marks)

Show Answers Only
  1. `$1439`
  2. `94\ \  text(months)`
Show Worked Solution

i.  `A_n = 200\ 000r^n – M(1 + r + r^2 + … + r^(n-1))`

`text(Find)\ M\ text(such that)\ A_n = 0\ text(when)\ n = 300`

`0 = 200\ 000(1.006^300) – M(1 + 1.006+ … + 1.006^299)`

`=>  text(Note that ) (1 + 1.006+ … + 1.006^299)\ \ text(is a)`

`text(GP)\ text(where)\ a = 1, r = 1.006, n = 300`

`0 = 200\ 000(1.006^300) – M ({1(1.006^300-1)}/(1.006 – 1))`

`M ({(1.006^300-1)}/(0.006))` `= 200\ 000(1.006^300)`
`M` `= (1\ 203\ 439.36…)/(836.199…)`
  `= 1439.177…`
  `= $1439\ \ text{(nearest dollar)}`

 
`:.\ text(The minimum monthly repayment is $1439.)`

 

ii.  `text(If)\ M = 2800,\ text(find)\ n\ text(when)\ A_n = 0`

`0 = 200\ 000(1.006^n) – 2800({(1.006^n – 1)}/0.006)`

`0 = 1200(1.006^n) – 2800(1.006^n) + 2800`

`0 = -1600(1.006^n) + 2800`

`1600(1.006^n)` `= 2800`
`1.006^n` `= 2800/1600`
`n xx ln 1.006` `= ln­ \ 1.75`
`n` `= (ln­ \ 1.75)/ln 1.006`
  `= 93.54…`
  `=94\ \ text{months  (nearest month)}`

 

`:.\ text(It will take Joe 94 months to repay the loan.)`

Financial Maths, 2ADV M1 2009 HSC 8b

One year ago Daniel borrowed $350 000 to buy a house. The interest rate was 9% per annum, compounded monthly. He agreed to repay the loan in 25 years with equal monthly repayments of $2937.

  1. Calculate how much Daniel owed after his first monthly repayment.    (1 mark)

Daniel has just made his 12th monthly repayment. He now owes $346 095. The interest rate now decreases to 6% per annum, compounded monthly.

 

The amount  `$A_n`, owing on the loan after the `n`th monthly repayment is now calculated using the formula  
 
`qquad qquad A_n=346,095xx1.005^n-1.005^(n-1)M-\ ... -1.005M-M`
  
where `$M` is the monthly repayment, and `n=1,2,\ ...,288`.   (DO NOT prove this formula.)

  1. Calculate the monthly repayment if the loan is to be repaid over the remaining 24 years (288 months).    (3 marks)

  2. Daniel chooses to keep his monthly repayments at $2937. Use the formula in part (ii) to calculate how long it will take him to repay the $346 095.   (3 marks)

  3. How much will Daniel save over the term of the loan by keeping his monthly repayments at $2937, rather than reducing his repayments to the amount calculated in part (ii)?   (1 mark)

Show Answers Only
  1. `$349\ 688`
  2. `$2270.31\ \ text{(nearest cent)}`
  3. `178.37\ text(months)`
  4. `$129\ 976.59\ \ text{(nearest cent)}`
Show Worked Solutions

i.   `text(Let)\ L_n= text(the amount owing after)\ n\ text(months)`

`text(Repayment)\ =M=$2937\ \ text(and)\ \ r=text(9%)/12=0.0075\ text(/month)`

`:.\ L_1` `=350\ 000(1+r)-M`
  `=350\ 000(1.0075)-2937`
  `=349\ 688`

 

`:.\ text(After 1 month, the amount owing is)\  $349\ 688`

 

ii.   `text(After 12 repayments, Daniel owes)\ $346\ 095,\  text(and)\ r darr 6%`

`:.\ r=(6%)/12=0.005`

`text(Loan is repaid over the next 24 years. i.e.)\ $A_n=0\ text(when)\  n=288`

`A_n` `=346\ 095(1.005^n)-1.005^(n-1)M-\ ..\ -1.005M-M`
  `=346\ 095(1.005^n)-M(1+1.005+..+1.005^(n-1))`
`A_288` `=346\ 095(1.005^288)-M(1+1.005+..+1.005^287)=0`

`=>\ GP\ text(where)\ a=1,\ text(and)\  r=1.005`

MARKER’S COMMENT: Careless setting out and poor handwriting, especially where indexes were involved, was a major contributor to errors in this question.

`M((1(1.005^288-1))/(1.005-1))=346\ 095(1.005^288)`

`M` `=(1\ 455\ 529.832)/641.1158`
  `=2270.31`

 

`:.\ text{Monthly repayment is $2270.31  (nearest cent)}` 

 

iii.  `text(Given)\ $M\ text(remains at $2937, find)\  n\ text(such that)`

`$A_n=0\ text{(i.e. loan fully paid off)}`

`:. 346\ 095(1.005^n)-2937((1(1.005^n-1))/(1.005-1))` `=0`
`346\ 095(1.005^n)-587\ 400(1.005^n-1)` `=0`
`(346\ 095-587\ 400)(1.005^n)+587\ 400` `=0`
♦♦ A poorly answered question.
MARKER’S COMMENT: Many students struggled to handle the exponential and logarithm calculations in this question.
ALGEBRA TIP: Dividing by `(1.005-1)` in part (iii) is equivalent to multiplying by 200, and cleans up working calculations (see Worked Solutions).
 

`241\ 305(1.005^n)` `=587\ 400`
`ln1.005^n` `=ln((587\ 400)/(241\ 305))`
`n` `=ln2.43426/ln1.005`
  `=178.37..`

 

`:.\ text{He will pay off the loan in 179 months (note the}`

`text{last payment will be a partial payment).}`

 

iv.  `text(Total paid at $2937 per month)`

`= 2937xx178.37=$523\ 872.69`

`text(Total paid at $2270.31 per month)`

`=2270.31xx288=$653,849.28`

 

`:.\ text(The amount saved)`

`=653\ 849.28-523\ 872.69`

`=$129\ 976.59\ \ text{(nearest cent)}`

Financial Maths, 2ADV M1 2012 HSC 15c

Ari takes out a loan of $360 000. The loan is to be repaid in equal monthly repayments, `$M`, at the end of each month, over 25 years (300 months). Reducible interest is charged at 6% per annum, calculated monthly.

Let  `$A_n`  be the amount owing after the `n`th repayment.

  1. Write down an expression for the amount owing after two months, `$A_2`.   (1 mark)

  2. Show that the monthly repayment is approximately $2319.50.   (2 marks)

  3. After how many months will the amount owing, `$A_n`, become less than $180 000.   (3 marks)

Show Answers Only
  1.  `$A_2=(360\ 000)(1.005^2)-M(1+1.005)`
  2. `text{Proof (See Worked Solutions)}`
  3. `202\ text(months)`
Show Worked Solutions
i.    `A_1` `=360\ 000(1+text(6%)/12)-M`
  `=360\ 000(1.005)-M`
`A_2` `=[360\ 000(1.005)-M](1.005)-M`
  `=360\ 000(1.005^2)-M(1.005)-M`
  `=360\ 000(1.005^2)-M(1+1.005)`

 

ii.  `A_n=360\ 000(1.005^n)-M(1+1.005^1+ … +1.005^(n-1))`

`text(When)\  n=300,\ A_n=0`

`0=360\ 000(1.005^300)-M(1+1.005^1+….+1.005^299)`

`360\ 000(1.005^300)` `=M((a(r^n-1))/(r-1))`
`M((1(1.005^300-1))/(1.005-1))` `=360\ 000(1.005^300)`
`:.M` `=((1\ 607\ 389.13)/692.994)`
  `~~2319.50\ \ \ text(… as required)`

 

iii.   `text(Find)\ n\ text(such that)\  $A_n<$180\ 000`

`360\ 000(1.005^n)-2319.50((1.005^n-1)/(1.005-1))` `<180\ 000`
`360\ 000(1.005^n)-463\ 900(1.005^n-1)` `<180\ 000`
`-103\ 900(1.005^n)+463,900` `<180\ 000`
♦ Mean mark 38%
MARKER’S COMMENT: Challenging calculations using logarithms are common in this topic. A high percentage of students consistently struggle in this area.
`103\ 900(1.005^n)` `>283\ 900`
`1.005^n` `>(283\ 900)/(103\ 900)`
`n(ln1.005)` `>ln((283\ 900)/(103\ 900))`
`n` `>1.005193/0.0049875`
`n` `>201.54`

 

`:.\ text(After 202 months,)\  $A_n< $180\ 000.`

Financial Maths, 2ADV M1 2013 HSC 13d

A family borrows $500 000 to buy a house. The loan is to be repaid in equal monthly instalments. The interest, which is charged at 6% per annum, is reducible and calculated monthly. The amount owing after  `n`  months, `$A_n`, is given by

`qquad qquadA_n=Pr^n-M(1+r+r^2+ \ .... +r^(n-1))\ \ \ \ \ \ \ \ \ ` (DO NOT prove this)

where  `$P`  is the amount borrowed, `r=1.005`  and  `$M`  is the monthly repayment.

  1. The loan is to be repaid over 30 years. Show that the monthly repayment is $2998 to the nearest dollar.     (2 marks)

  2. Show that the balance owing after 20 years is $270 000 to the nearest thousand dollars.             (1 mark)

After 20 years the family borrows an extra amount, so that the family then owes a total of $370 000. The monthly repayment remains $2998, and the interest rate remains the same.

  1. How long will it take to repay the $370 000?     (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `text(193 months)`
Show Worked Solutions

i.    `text(Find)\  $M\  text(such that the loan is repaid over 30 years.)`

`n=30xx12=360\ text(periods)\ \ \   r=1+6/12%=1.005`

`A_360` `=500\ 000 (1.005^360)-M(1+1.005+..+1.005^359)=0`

`=>GP\ text(where)\ a=1,\ \ r=1.005,\ \ \ n=360`

`M((1(1.005^360-1))/(1.005-1))` `=500\ 000(1.005^360)`
`M(1004.515)` `=3\ 011\ 287.61`
`M` `=2997.75`

 

`:.$M=$2998\ \ text{(nearest dollar) … as required}`

 

 ii.    `text(Find)\  $A_n\ text(after 20 years)\ \ \ =>n=20xx12=240` 

`A_240` `=500\ 000(1.005^240)-2998(1+1.005+..+1.005^239)`
  `=1\ 655\ 102.24-2998((1(1.005^240-1))/(1.005-1))`
  `=269\ 903.63`
  `=270\ 000\ \ text{(nearest thousand) … as required}`
MARKER’S COMMENT: Within the GP formula, many students incorrectly wrote the last term as `1.005^240` rather than `1.005^239`. Note `T_n=ar^(n-1)`.

 

 

iii.  `text(Loan)=$370\ 000`

`text(Find)\  n\  text(such that)\  $A_n=0,\ \ \ M=$2998`

`A_n=370\ 000(1.005^n)-2998(1+1.005+..+1.005^(n-1))=0`

♦♦ Mean mark 33%
COMMENT: Another good examination of working with logarithms. Students should understand why they must ’round up’ their answer in this question.
`370\ 000(1.005^n)` `=2998((1(1.005^n-1))/(1.005-1))` 
`370\ 000(1.005^n)` `=599\ 600(1.005^n-1)`
`229\ 600(1.005^n)` `=599\ 600`
`ln1.005^n` `=ln((599\ 600)/(229\ 600))`
`n` `=ln2.6115/ln1.005`
`n` `=192.4`

 
`:.\ text(The loan will be repaid after 193 months.)`