Application of Geometric Sequence and Series

In this tutorial we discuss about 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 be the value of an automobile after three years hence which is now purchased for Dollar 4500.

Solution:
            Let {a_1} = 4500 = purchased value of the automobile.
            The value of automobile at the end of 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 automobile at the end of 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 third year
                         = {a_1}{\left( {0.85} \right)^3}
            Hence, the value of the automobile at the end of 3years hence now purchased for 4500
                         = 4500{\left( {0.85} \right)^3} = 4500\left(  {0.6141} \right) = 2763.45