Application of Geometric Sequence and Series

In this tutorial we discuss the related problems of application of geometric sequence and geometric series.

Example:                                                       

A line is divided into six parts forming a geometric sequence. If the shortest length is 3cm and the longest is 96cm, find the length of the whole line.

Solution:
Given that
{a_1} = 3, {a_n} = 96, n = 6

Since   {a_n} = {a_1}{r^{n - 1}}

we have
96 = 3 \times {r^5}          \Rightarrow r = 2

For the length of the whole line, we have
{S_n} = \frac{{{a_1}\left( {{r^n} - 1} \right)}}{{r - 1}} (As r > 1)
{S_n} = \frac{{3\left( {{2^6} - 1} \right)}}{{2 - 1}} = 3\left( {64 - 1} \right) = 189cm

Therefore, the whole length of the line is equal to 189cm.

Example:

The value of an automobile depreciates at the rate of 15\% per year. What will the value of an automobile be after three years if it is purchased for 4500 dollars?

Solution:
Let {a_1} = 4500 = purchased value of the automobile.

The value of automobile at the end of the first year
 = {a_1} - {a_1}\left( {\frac{{15}}{{100}}} \right) = {a_1}\left( {1 - 0.05} \right) = {a_1}\left( {0.85} \right)

The value of the automobile at the end of the second year
 = {a_1}\left( {0.85} \right) - {a_1}\left( {0.85} \right)\left( {\frac{{15}}{{85}}} \right) = {a_1}\left( {0.85} \right)\left[ {1 - \frac{{15}}{{100}}} \right]
 = {a_1}\left( {0.85} \right)\left( {1 - 0.85} \right) = {a_1}\left( {0.85} \right)\left( {0.85} \right) = {a_1}{\left( {0.85} \right)^2}

Similarly, the value at the end of the third year
 = {a_1}{\left( {0.85} \right)^3}

Hence, the value of the automobile at the end of 3 years after being purchased for 4500
 = 4500{\left( {0.85} \right)^3} = 4500\left( {0.6141} \right) = 2763.45