First month :

After paying $300, amount outstanding
= $15000 - $300
= $14700

Now 0.5% means 0.5/100 or 1/200

Together with the interest incurred, total outstanding amount at the end of first month

= $14700 × 100.5/100
= $14700 × 1.005
= $14773.50


Second month :

After paying $300, amount outstanding
= $14773.50 - $300
= $14473.50

Together with the interest incurred, total outstanding amount at the end of second month

= $14473.50 × 1.005
= $14545.8675
= $14545.87 (nearest cent)

(shown)

We notice a pattern :

For the first month,
Outstanding amount at the end of the month
= $(15000 - 300)(1.005)
= $15000(1.005) - $300(1.005)

For the second month,
Outstanding amount at the end of the month

= [($15000(1.005) - $300(1.005) - $300](1.005)

= $15000(1.005²) - $300(1.005²) - $300(1.005)


3rd month :

[$15000(1.005²) - $300(1.005²) - $300(1.005) - $300](1.005)

= $15000(1.005³) - $300(1.005³) - $300(1.005²) - $300(1.005)



So for the nth month , it would be :

$15000(1.005ⁿ) - $300(1.005ⁿ) - $300(1.005ⁿ-¹) - ... - $300(1.005)


= $15000(1.005ⁿ) - $300(1.005ⁿ + 1.005ⁿ-¹ + ... + 1.005)



Notice the back terms form the sum of a G.P with first term 1.005 and common ratio 1.005. There are n terms in total.

Since the common ratio > 1, we use the sum formula a(rⁿ - 1)/(r - 1)


So amount owed at the end of the nth month


= $15000(1.005ⁿ) - $300(1.005(1.005ⁿ - 1)/(1.005 - 1))


= $15000(1.005ⁿ) - $300(1.005(1.005ⁿ - 1)/0.005)

= $15000(1.005ⁿ) - $60300(1.005ⁿ - 1)

= $15000(1.005ⁿ) - $60300(1.005ⁿ) + $60300

= $[60300 - 45300(1.005ⁿ)]

(Shown)

If he incurred no interest, he will take :

$15000 ÷ $300 per month
= 50 months


With interest, we look for the value of n when the outstanding amount is $0.


[60300 - 45300(1.005ⁿ)] = 0


60300 = 45300(1.005ⁿ)

1.005ⁿ = 60300/45300 = 201/151


n = log1.005 (201/151)

n = ln(201/151) / ln1.005

n ≈ 57.3479079


Least n required = 58 , so 58 months are needed.


58 - 50 = 7


8 more months are needed

Edited.