# Geometric Series

Another known series is the geometric series also called as Geometric progression. The following term of the series is obtained by multiplying a constant multiplier called a common ratio, equivalent to the common difference of arithmetic progression. It is also an advantage that we are equipped with our knowledge in other topic especially evaluating exponential equations. That skill will help us solve harder problem about this type of series.

Formula derivation:

Nth term                                             Formula

1st term                                                 $a_1$

2nd term                                               $a_1\cdot r$

3rd term                                                $a_1\cdot r^2$

4th term                                                $a_1\cdot r^3$

5th term                                                 $a_1\cdot r^4$

Now, observe how the nth term of the series and the power of $r$ related with each other.

Without going further to 6th term, 7th term to nth term we can solve it using the following formula.

$a_n=a_1\cdot r^{n-1}$

where $a_n$– is the nth term, $a_1$– is the first term, $n$– is the number of terms.

Worked Problem 1:

Find the 8th term of the series {4,8,16…}.

Solution:

Using the our formula,

$a_n=a_1\cdot r^{n-1}$ $a_8=4\cdot 2^{8-1}$ $a_8=4\cdot 2^7$ $a_8=512$

Worked Problem 2:

The 2nd term of a geometric progression is 8 and the 7th term is $\displaystyle\frac{1}{4}$. What is the first term?

Solution:

$a_n=a_1\cdot r^{n-1}$ can be written as $a_n=a_m\cdot r^{n-m}$

$a_7=a_2\cdot r^{7-2}$ $\displaystyle\frac{1}{4}=8\cdot r^5$ $\displaystyle\frac{1}{4\cdot 8}=r^5$ $\displaystyle\frac{1}{32}=r^5$ $r=\displaystyle\frac{1}{2}$

Since the second term is 8 the previous number is the first term and can be obtained by multiplying the second term by 2. Making 16 is the first term.

Worked Problem 3:

The nth term of geometric progression is given by the equation $a_n=3^{n-1}$. Find the 5th term.

Solution:

$a_n=3^{n-1}$ $a_5=3^{5-1}$ $a_5=81$

Worked Problem 4:

The terms 4,x,25 forms a geometric sequence. What is the value of x?

Solution:

Let r be the common ration.

$4r_1=x$, also $xr_2=25$

$r_1=r_2$ $\displaystyle\frac{x}{4}=\displaystyle\frac{25}{x}$ $x^2=100$ $x=10$

By the way, x is also called the geometric mean.

$GM=\sqrt{ab}$ $GM=\sqrt{4\cdot 25}$ $GM=\sqrt{100}$ $GM=10$