# Find the indicated nth term of the geometric sequence. 6th term: a_5 = {4} / {81}, a_8 = {4} /...

## Question:

Find the indicated nth term of the geometric sequence.

6th term: {eq}\displaystyle a_5 = \dfrac {4}{81},\ a_8 = \dfrac{4} {2187} {/eq}

## Geometric Progressions

A geometric progression is a type of progression in which the ratio of subsequent terms is a constant. Using this definition we can define the {eq}n^{th} {/eq} term of the geometric progression.

The {eq}n^{th} {/eq} term of a Geometric Progression can be written down as -

{eq}a_n = ar^{n-1} {/eq}

Here, we are given that -

{eq}a_5 = \dfrac {4}{81}\\ a_8 = \dfrac{4}{2187} {/eq}

Thus we can find that -

{eq}\dfrac{a_8}{a_5} = \dfrac{ar^7}{ar^4} = \dfrac{\dfrac{4}{2187}}{\dfrac {4}{81}}\\ r^3 = \dfrac{81}{2187}\\ r^3 = \dfrac{81}{2187}\\ r^3 = \dfrac{1}{27}\\ r = \dfrac{1}{3} {/eq}

Thus we can find the {eq}6^{th} {/eq} term by multiplying the {eq}5^{th} {/eq} term by the common ratio.

The {eq}6^{th} {/eq} term is -

{eq}\dfrac {4}{81} \times \dfrac{1}{3} = \dfrac {4}{243} {/eq}

Thus the {eq}6^{th} {/eq} term is - {eq}\dfrac {4}{243} {/eq}