# 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 $$3$$cm and the longest is $$96$$cm, 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 $$r > 1$$)

$${S_n} = \frac{{3\left( {{2^6} – 1} \right)}}{{2 – 1}} = 3\left( {64 – 1} \right) = 189$$cm

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

__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 $$4500$$

$$ = 4500{\left( {0.85} \right)^3} = 4500\left( {0.6141} \right) = 2763.45$$