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