A sequence of numbers \(\left\{ {{a_n}} \right\}\) is called a geometric sequence if the quotient of successive terms is a constant, called the common ratio. Thus \({\large\frac{{{a_{n + 1}}}}{{{a_n}}}\normalsize} = q\) or \({a_{n + 1}} = q{a_n}\) for all terms of the sequence. It’s supposed that \(q \ne 0\) and \(q \ne 1.\)

For any geometric sequence:

\[{a_n} = {a_1}{q^{n – 1}}.\]

A geometric series is the indicated sum of the terms of a geometric sequence. For a geometric series with \(q \ne 1,\)

\[

{{S_n} }={ {a_1} + {a_2} + \ldots + {a_n} }={ {a_1}\frac{{1 – {q^n}}}{{1 – q}},\;\;}\kern-0.3pt

{q \ne 1.}

\]

We say that the geometric series converges if the limit \(\lim\limits_{n \to \infty } {S_n}\) exists and is finite.Otherwise the series is said to diverge.

Let \(S = \sum\limits_{n = 0}^\infty {{a_n}} \) \(= {a_1}\sum\limits_{n = 0}^\infty {{q^n}} \) be a geometric series. Then the series converges to \(\large\frac{{{a_1}}}{{1 – q}}\normalsize\) if \(\left| q \right| \lt 1,\) and the series diverges if \(\left| q \right| \gt 1.\)

## Solved Problems

Click a problem to see the solution.

### Example 1

Find the sum of the first \(8\) terms of the geometric sequence \(3,6,12, \ldots \)### Example 2

Find the sum of the series \(1 – 0,37 + 0,{37^2} \) \(- 0,{37^3} + \ldots \)### Example 3

Find the sum of the series### Example 4

Express the repeating decimal \(0,131313 \ldots \) as a rational number.### Example 5

Show that### Example 6

Solve the equation### Example 7

The second term of an infinite geometric progression (\(\left| q \right| \lt 1\)) is \(21\) and the sum of the progression is \(112.\) Determine the first term and ratio of the progression.### Example 1.

Find the sum of the first \(8\) terms of the geometric sequence \(3,6,12, \ldots \)Solution.

Here \({a_1} = 3\) and \(q = 2.\) For \(n = 8\) we have

\[

{{S_8} = {a_1}\frac{{1 – {q^8}}}{{1 – q}} }

= {3 \cdot \frac{{1 – {2^8}}}{{1 – 2}} }

= {3 \cdot \frac{{1 – 256}}{{\left( { – 1} \right)}} }={ 765.}

\]

### Example 2.

Find the sum of the series \(1 – 0,37 + 0,{37^2} \) \(- 0,{37^3} + \ldots \)Solution.

This is an infinite geometric series with ratio \(q = -0,37.\) Hence, the series converges to

\[

{S = \sum\limits_{n = 0}^\infty {{q^n}} }

= {\frac{1}{{1 – \left( { – 0,37} \right)}} }

= {\frac{1}{{1 + 0,37}} }

= {\frac{1}{{1,37}} }

= {\frac{{100}}{{137}}.}

\]