Math, asked by ducko13, 1 month ago

The value of an antique increases at a rate of 3.5% every year. In 2000, the antique was purchased for $5000.
a.) Determine an explicit formula to represent the value of the antique since year 2000.
b.) Use your formula to write the first three terms of the sequence.
c.) What is the value of the antique in 2008?
d.) In which year will the value of the antique be $20489?

Answers

Answered by tennetiraj86
2

Step-by-step explanation:

Given :-

The value of an antique increases at a rate of 3.5% every year. In 2000, the antique was purchased for $5000.

To find :-

Find the following :-

a.) Determine an explicit formula to represent the value of the antique since year 2000.

b.) Use your formula to write the first three terms of the sequence.

c.) What is the value of the antique in 2008?

d.) In which year will the value of the antique be $20489?

Solution :-

(a)

The Cost Price of an antique in the year 2000= $5000

Increasing percentage of the antique every year = 3.5%

Increasing value = 3.5% of Cost Price

=> 3.5% of 5000

= (3.5/100)×5000

= 3.5×50

= 35×5

= 175

Increasing value = 175

The value of the antique in the year 2001

=5000+175

=$ 5175

The Increasing value after 1 year

= 3.5% of 5175

=> (3.5/100)×5175

=> (35/1000)×5175

= 181.125

The cost price of the antique in 2002

= 5175 + 181.125

= $ 5356.125

We have ,

5000,5175 , 5356.125,...

First term = 5000

Common ratio

= 5175/5000 = 1.035

= 5356.125/5175 = 1.035

Since the Common ratio is same throughout the series

They are in the GP

We know that

The general form of a GP

= an = a×r^(n-1)

=>an = 5000×(1.035)^(n-1)

The Formula = 5000×(1.035)^(n-1)-------(1)

(b)

Put n = 2 then

a2 = 5000×(1.035)^(2-1)

=> a2 = 5000×1.035

=> a2 = 5175

Put n = 3 then

a3 = 5000×(1.035)^(3-1)

=>a3 = 5000×(1.035)²

=> a3 = 5000×1.035×1.035

=> a3 = 5356.125

The first three terms of the sequence are 5000, 5175,5356.125

(c)

Given year = 2008

The initial value in 2000 = $5000

Number of years = 2008-2000+1 = 9

Put n = 9 in (1) then

a9 = 5000×(1.035)^(9-1)

=> a9 = 5000×(1.035)⁸

=> a9 = 5000×1.3168090369634

=> a9 = 6584.0451848170

=> a9 = 6584.045

The value of the antique in the year 2008 is

The value of the antique in the year 2008 is $ 6584.045

(d)

Given cost of the antique = $20489

Let an = $20489

=> a×r^(n-1) = $20489

=> 5000×(1.035)^(n-1) = 20489

=> (1.035)^(n-1) = 20489/5000

=> (1.035)^(n-1) = 4.0978

=> (1.035)^(n-1) = (1.035)⁴¹

Since the bases are equal then exponents must be equal

=> n-1 = 41

=> n = 41+1

=> n = 42

2000 year + 42 years= 2041 year

In 2041 the Cost of the antique is

$ 20489

Answer:-

(a)an explicit formula to represent the value of the antique since year 2000 is 5000×(1.035)^(n-1)

(b)The first three terms of the sequence are 5000, 5175,5356.125

(c) The value of the antique in the year 2008 is $ 6584.045

(d) In 2041 the Cost of the antique is

$ 20489

Used formulae:-

  • The general term of a GP is a×r^(n-1)
  • a = first term
  • r = Common ratio
  • n = number of terms
Similar questions