FIND BALANCE ON INVESTMENT AFTER N YEARS INVOLVING COMPOUND INTEREST

Example 1 :

An investor deposits $70 on the first day of each month into an account that pays 2% interest per annum compounded monthly. What is the balance of this annuity at the end of 4 years?

Solution :

We need to consider each of the 48 deposits separately.

The first deposit gains interest for 48 months, the second month deposit gains interest for 47 months and so on.

So the last month deposit (48 month's deposit) will give us only 1 month interest.

Principal  =  $70, Rate of interest (r)  =  2%  =  0.02

Interest is compounding monthly

1 year  =  12 months 

4 years  =  48 months

A  =  P(1 + r/n)^nt

The first deposit is gaining interest for 48 months.

A  =  70 [1 + (0.02/12)]⁴⁸

A  =  70 [1 + (0.02/12)]⁴⁸

The second deposit is gaining interest for 47 months.

A  =  70 [1 + (0.02/12)]⁴⁷

A  =  70 [1 + (0.02/12)]⁴⁷

.....

......

The last deposit is gaining interest for 1 month.

A  =  70 [1 + (0.02/12)]¹

A  =  70 [1 + (0.02/12)]¹

By writing these amounts as sequence we get

70 [1 + (0.02/12)]¹_, 70 [1 + (0.02/12)]²_, 70 [1 + (0.02/12)]³_,^.........   70 [1 + (0.02/12)]⁴⁸

This is a geometric sequence having the common ratio [1 + (0.02/12)]

a  =  70 [1 + (0.02/12)] and r  =  [1 + (0.02/12)]

s_n  =  a(1 - rⁿ)/(1 - r)

Example 2 :

Mrs. Bright has decided to retire as soon as she has set aside $1 million if she invests $10,000 in an account that earns 8.3% interest compound monthly 

(a)  How much money will she have in 20 years?

(b)  How long will it be before she can retire?

Solution :

(a)

A  =  P(1 + r/n)^nt

Principal  =  $10,000, r  =  8.3%

n  =  12(Compounding monthly), t  =  20 years

A₂₀  =  10000(1 + 0.083/12)¹²⁽²⁰⁾

A₂₀  =  52293.42

(b)  To find the time consumption to reach the cost $1 million

1000000  =  10000(1 + 0.083/12)^12t

100  =  (1.0069)^12t

log 100  =  12t log(1.0069)

2  =  12t(0.0029)

689.65  =  12t

t  =  57 years

Example 3 :

An investor deposits $100 on the first day of each month in an account pays 1.8% interest compounded monthly. Find the balance in the account after 5 years.

Solution :

The investor is going to deposit $100 on each month.

1 year  =  12 months

5 years  =  60 months

The first month deposit is going to earn 60 months interest. If we calculate the value of 100 that is invested today after 60 months, we should use the formula given below.

A  =  P(1 + r/n)^nt

The amount after 60 months after depositing $100 for the first month.

A  =  100(1 + 0.018/12)⁶⁰

The amount after 59 months after depositing $100 for the second month.

A  =  100(1 + 0.018/12)⁵⁹

..............

A  =  100(1 + 0.018/12)¹

Total balance  =  100(1 + 0.018/12)¹ + 100(1 + 0.018/12)² + 

................. + 100(1 + 0.018/12)⁶⁰

a  =  100(1 + 0.018/12), r  =  (1 + 0.018/12) and n  =  60

s_n  =  a(1 - rⁿ)/(1 - r)

s_n  =  100(1 + 0.018/12)[1 - (1 + 0.018/12)⁶⁰)/(1 - (1+0.018/12)]

  =  $6282.78

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