FINANCIAL APPLICATIONS OF GEOMETRIC SEQUENCES AND SERIES

Example 1 : 

A savings account has an annual interest rate of 3% compounded monthly. Find the amount in the account 18 months after $5000 is invested.

Solution : 

It is given that the annual interest rate is 3%.

The monthly interest rate : 

= 3/12

= 0.25%

The decimal form of 0.25% : 

= 0.25/100

= 0.0025

An increase of 0.25% is equivalent to multiplying by the factor (1 + 0.0025) or 1.0025. 

Amount in the account after 

1 month : 5000(1.0025)

2 months : 5000(1.0025)²

3 months : 5000(1.0025)³

This is a geometric sequence with 

a₁ = 5000(1.0025) and r = 1.0025

Use the formula a_n = a₁r^n-1 to calculate the amount in the account after 18 months. 

a₁₈ = 5000(1.0025)(1.0025)^{18 - 1}

= 5000(1.0025)(1.0025)¹⁷

= 5000(1.0025)¹⁸

= 5229.85

Example 2 : 

An investment account offers 2% annual interest. Neil invests $1000 in this account for 5 years. During those 5 years there is inflation of 2.5% per annum. What is the value in real terms of Neil’s investment after those 5 years.

Solution : 

Annual interest rate (c) : 2%

Annual inflation rate (i) = 2.5%

The annual real terms percentage change is 

= c - i

= 2 - 2.5

= -0.5%

A decrease of 0.5% is equivalent to multiplying by the factor (1 - 0.005) or 0.995. 

The value in real terms after 

1 year : 1000(0.995)

2 years : 1000(0.995)²

3 years : 1000(0.995)³

This is a geometric sequence with 

a₁ = 1000(0.995) and r = 0.995

Use the formula a_n = a₁r^n-1 to find the value of the investment in real terms after 5 years. 

a₅ = 1000(0.995)(0.995)^{5 - 1}

= 1000(0.995)(0.995)⁴

= 1000(0.995)⁵

≈ $975

Example 3 :

After how many complete years will a starting capital of $5000 first exceed 10000 if it grows at 6% per annum.

Solution : 

a = 5000

r = 6%

a (1 + r)ⁿ > 10000

5000 (1 + 0.06)ⁿ > 10000

5000 (1.06)ⁿ > 10000

(1.06)ⁿ > 10000/5000

(1.06)ⁿ > 2

n ln 1.06 > ln 2

n > ln 2/ln 1.06

n > 0.693/0.058

n > 11.94

So, approximately 12 years.

Miles joined the Apollo Golf and Gym Club in January 2005. The annual fee was £672 then. Since 2005 the fees have increased steadily by 4% every year.

(a) How much was the membership fee in 2010?

(b) How much in total did Miles pay in membership fees from 2005 to 2010?

Solution :

Annual fee = 672

Since the fee increases steadily 4%, it comes under geometric sequence.

672, 104% of 672, 104% of (104% of 672),...........

672, 1.04(672), (104)² 672, ..................

a = 672, r = 1.04

a)

To find the membership fee in 2010, we have to find the 5th term of the sequence.

a_n = a₁r^n-1

a₅ = 672(1.04)^5-1

= 672(1.04)⁴

= 672(1.169)

= 786.14

b) s_n = a(rⁿ - 1)/(r - 1)

= 672(1.04⁵ - 1) / (1.04 - 1)

= 672(1.2166 - 1) / 3

= 20437.2

So, the total fee from 2005 to 2010 is 20437.2.

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