1.5 Finite geometric series | Sequences and series

1.5 Finite geometric series (EMCDZ)

When we sum a known number of terms in a geometric sequence, we get a finite geometric series. We generate a geometric sequence using the general form:

${T}_{n} = a \cdot {r}^{n-1}$

where

$n$ is the position of the sequence;
${T}_{n}$ is the $n$ $^{\text{th}}$ term of the sequence;
$a$ is the first term;
$r$ is the constant ratio.

General formula for a finite geometric series (EMCF2)

$\begin{align*} {S}_{n} &= a+ ar+ a{r}^{2}+\cdots + a{r}^{n-2} + a{r}^{n-1} \ldots (1) \\ r \times {S}_{n} &= \qquad ar+ a{r}^{2}+\cdots + a{r}^{n-2} + a{r}^{n-1} + ar^{n} \ldots \ldots (2) \\ \text{Subtract eqn. } (2) &\text{ from eqn. } (1) \\ \therefore {S}_{n} - r{S}_{n} &= a + 0 + 0 + \cdots - ar^{n} \\ {S}_{n} - r{S}_{n} &= a - ar^{n} \\ {S}_{n}(1 - r) &= a(1 - r^{n}) \\ \therefore {S}_{n} &= \frac{a(1 - r^{n})}{1 - r} \quad (\text{where } r \ne 1) \end{align*}$

The general formula for determining the sum of a geometric series is given by:

${S}_{n} = \frac{a(1 - r^{n})}{1 - r} \qquad \text{where } r \ne 1$

This formula is easier to use when $r < 1$ .

Alternative formula:

$\begin{align*} {S}_{n} &= a+ ar+ a{r}^{2}+\cdots + a{r}^{n-2} + a{r}^{n-1} \ldots \ldots (1) \\ r \times {S}_{n} &= \qquad ar+ a{r}^{2}+\cdots + a{r}^{n-2} + a{r}^{n-1} + ar^{n} \ldots \ldots (2) \\ \text{Subtract eqn. } (1) &\text{ from eqn. } (2) \\ \therefore r{S}_{n} - {S}_{n} &= ar^{n} - a \\ {S}_{n}(r - 1) &= a(r^{n}-1) \\ \therefore {S}_{n} &= \frac{a(r^{n}-1)}{r - 1} \quad (\text{ where } r \ne 1) \end{align*}$

The general formula for determining the sum of a geometric series is given by:

${S}_{n} = \frac{a(r^{n}-1)}{r - 1} \qquad \text{where } r \ne 1$

This formula is easier to use when $r > 1$ .

Video: 2875

Worked example 11: Sum of a geometric series

Calculate: $\sum _{k = 1}^{6}{32 \left( \frac{1}{2} \right)^{k-1}}$

Write down the first three terms of the series

$\begin{align*} k=1; \quad T_{1}&= {32 \left( \frac{1}{2} \right)^{0}} = 32 \\ k=2; \quad T_{2}&= {32 \left( \frac{1}{2} \right)^{2-1}} = 16 \\ k=3; \quad T_{3}&= {32 \left( \frac{1}{2} \right)^{3-1}} = 8 \end{align*}$

We have generated the series $32 + 16 + 8 + \cdots$

Determine the values of $a$ and $r$

$\begin{align*} a &= T_{1} = 32 \\ r &= \frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{2}} = \frac{1}{2} \end{align*}$

Use the general formula to find the sum of the series

$\begin{align*} {S}_{n} &= \frac{a(1 - r^{n})}{1 - r}\\ {S}_{6} &= \frac{32(1 - \left( \frac{1}{2} \right)^{6})}{1 - \frac{1}{2}} \\ &= \frac{32\left(1 - \frac{1}{64} \right)}{\frac{1}{2}} \\ &= 2 \times 32 \left( \frac{63}{64} \right) \\ &= 64 \left( \frac{63}{64} \right) \\ &= 63 \end{align*}$

Write the final answer

$\sum _{k = 1}^{6}{32 \left( \frac{1}{2} \right)^{k-1}} = 63$

Worked example 12: Sum of a geometric series

Given a geometric series with $T_{1} = -4$ and $T_{4} = 32$ . Determine the values of $r$ and $n$ if $S_{n} = 84$ .

Determine the values of $a$ and $r$

$\begin{align*} a &= T_{1} = -4 \\ T_{4} &= ar^{3} = 32 \\ \therefore -4 r^{3} &= 32 \\ r^{3} &= -8 \\ \therefore r &= -2 \end{align*}$

Therefore the geometric series is $-4 + 8 -16 + 32 \ldots$ Notice that the signs of the terms alternate because $r < 0$ .

We write the general term for this series as $T_{n} = -4(-2)^{n-1}$ .

Use the general formula for the sum of a geometric series to determine the value of $n$

$\begin{align*} {S}_{n} &= \frac{a(1 - r^{n})}{1 - r}\\ \therefore 84 &= \frac{-4(1 - (-2)^{n})}{1 - (-2)} \\ 84 &= \frac{-4(1 - (-2)^{n})}{3} \\ - \frac{3}{4} \times 84 &= 1 - (-2)^{n} \\ - 63 &= 1 - (-2)^{n} \\ (-2)^{n} &= 64 \\ (-2)^{n} &= (-2)^{6} \\ \therefore n &= 6 \end{align*}$

Write the final answer

$r = -2 \text{ and } n = 6$

Worked example 13: Sum of a geometric series

Use the general formula for the sum of a geometric series to determine $k$ if $\sum _{n = 1}^{8}{k \left( \frac{1}{2} \right)^{n}} = \frac{255}{64 }$

Write down the first three terms of the series

$\begin{align*} n=1; \quad T_{1}&= {k \left( \frac{1}{2} \right)^{1}} = \frac{1}{2}k \\ n=2; \quad T_{2}&= {k \left( \frac{1}{2} \right)^{2}} = \frac{1}{4}k \\ n=3; \quad T_{3}&= {k \left( \frac{1}{2} \right)^{3}} = \frac{1}{8}k \end{align*}$

We have generated the series $\frac{1}{2}k + \frac{1}{4}k + \frac{1}{8}k + \cdots$

We can take out the common factor $k$ and write the series as: $k \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \right)$

$\therefore k \sum _{n = 1}^{8}{\left( \frac{1}{2} \right)^{n}} = \frac{255}{64}$

Determine the values of $a$ and $r$

$\begin{align*} a &= T_{1} = \frac{1}{2} \\ r &= \frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{2}} = \frac{1}{2} \end{align*}$

Calculate the sum of the first eight terms of the geometric series

$\begin{align*} \therefore {S}_{n} &= \frac{a(1 - r^{n})}{1 - r}\\ {S}_{8} &= \frac{\frac{1}{2}(1 - \left( \frac{1}{2} \right)^{8})}{1 - \frac{1}{2}}\\ &= \frac{\frac{1}{2}(1 - \left( \frac{1}{2} \right)^{8})}{\frac{1}{2}}\\ &= 1 - \frac{1}{256} \\ &= \frac{255}{256} \\ & \\ \therefore \sum _{n = 1}^{8}{\left( \frac{1}{2} \right)^{n}} &= \frac{255}{256} \end{align*}$

So then we can write:

$\begin{align*} k \sum _{n = 1}^{8}{\left( \frac{1}{2} \right)^{n}} &= \frac{255}{64} \\ k \left( \frac{255}{256} \right) &= \frac{255}{64} \\ \therefore k &= \frac{255}{64} \times \frac{256}{255} \\ &= \frac{256}{64} \\ &= 4 \end{align*}$

Write the final answer

$k = 4$

Sum of a geometric series

Textbook Exercise 1.9

Prove that $a + ar + a{r}^{2} + \cdots + a{r}^{n-1} = \frac{a(r^{n} - 1) }{r - 1}$ and state any restrictions.

\begin{align*} {S}_{n} &= a+ ar+ a{r}^{2}+\cdots + a{r}^{n-2} + a{r}^{n-1} \ldots (1) \\ r \times {S}_{n} &= \qquad ar+ a{r}^{2}+\cdots + a{r}^{n-2} + a{r}^{n-1} + ar^{n} \ldots (2) \\ \text{Subtract eqn. } (1) &\text{ from eqn. } (2) \\ \therefore r{S}_{n} - {S}_{n} &= ar^{n} - a \\ {S}_{n}(r - 1) &= a(r^{n} - 1) \\ \therefore {S}_{n} &= \frac{a(r^{n} - 1) }{r - 1} \end{align*}

where $r \ne 1$.

The eighth term of the sequence.

\begin{align*} a &= 1 \\ r &= \frac{T_{2}}{T_{1}} = -3 \\ T_{n} &= ar^{n-1} \\ \therefore T_{8} &= (1)(-3)^{8-1} \\ &= (1)(-3)^{7} \\ &= -2187 \end{align*}

The sum of the first eight terms of the sequence.

\begin{align*} S_{n} &= \frac{a(1-r^{n})}{1 - r}\\ \therefore S_{8} &= \frac{(1)(1-(-3)^{8})}{1 - (-3)}\\ &= \frac{1 - 6561}{4}\\ &= -\frac{6560}{4}\\ &= -1640 \end{align*}

Determine:

\[\sum _{n=1}^{4}3 \cdot {2}^{n-1}\]

\begin{align*} S_{4} &= 3 + 6 + 12 + 24 \\ &= 45 \end{align*}

Find the sum of the first $\text{11}$ terms of the geometric series $6+3+\frac{3}{2}+\frac{3}{4}+ \cdots$

\begin{align*} a &= 6 \\ r &= \frac{1}{2} \\ S_{n} &= \frac{a(1 - r^{n})}{1 - r} \\ S_{11} &= \frac{6(1 - \left( \frac{1}{2} \right)^{11})}{1 - \left( \frac{1}{2} \right)} \\ &= 12 \left( 1 - \frac{1}{2048} \right) \\ &= 12 \left( \frac{2047}{2048} \right) \\ &= \frac{6141}{512} \end{align*}

Show that the sum of the first $n$ terms of the geometric series $54+18+6+\cdots +5 {\left(\frac{1}{3}\right)}^{n-1}$ is given by $\left( 81-{3}^{4-n} \right)$.

\begin{align*} a &= 54 \\ r &= \frac{1}{3} \\ S_{n} &= \frac{a(1 - r^{n})}{1 - r} \\ &= \frac{54(1 - \left( \frac{1}{3} \right)^{n})}{1 - \left( \frac{1}{3} \right)} \\ &= \frac{54(1 - \left( \frac{1}{3} \right)^{n})}{ \frac{2}{3} } \\ &= 81 ( 1 - 3^{-n}) \\ &= 81 - 81 \cdot 3^{-n} \\ &= 81 - (3^{4} \cdot 3^{-n}) \\ &= 81 - 3^{4-n} \end{align*}

The eighth term of a geometric sequence is $\text{640}$. The third term is $\text{20}$. Find the sum of the first $\text{7}$ terms.

\begin{align*} T_{8} &= 640 = ar^{7} \\ T_{3} &= 20 = ar^{2} \\ \therefore \frac{T_{8}}{T_{3}} &= \frac{640}{20} \\ \frac{640}{20} &= \frac{ar^{7}}{ar^{2}} \\ 32 &= r^{5} \\ \therefore 2 &= r \\ \text{And } 20 &= ar^{2} \\ 20 &= a(2)^{2} \\ \frac{20}{4} &= a \\ \therefore 5 &= a \\ r &= 2 \\ S_{n} &= \frac{a(r^{n} - 1)}{r - 1} \\ S_{7} &= \frac{5((2)^{7} - 1)}{2 - 1} \\ &= 5(128 - 1) \\ &= 635 \end{align*}

Find the first three terms in the series.

\begin{align*} t = 1: \quad T_{1} &= 4 \\ t = 2: \quad T_{2} &= 2 \\ t = 3: \quad T_{3} &= 1 \\ 4; & 2; 1 \end{align*}

Calculate the number of terms in the series if $S_{n}=7\frac{63}{64}$.

\begin{align*} a &= 4 \\ r &= \frac{1}{2} \\ S_{n} &= \frac{a(1 - r^{n})}{1 - r} \\ \frac{511}{64} &= \frac{4(1 - \left( \frac{1}{2} \right)^{n})}{1 - \left( \frac{1}{2} \right)} \\ &= \frac{4 - 4\left( \frac{1}{2} \right)^{n}}{\frac{1}{2}} \\ \frac{511}{128} &= 4 - (2^{2} \cdot 2^{-n}) \\ 2^{2 - n}&= 4 - \frac{511}{128} \\ 2^{2 - n}&= \frac{1}{128} \\ 2^{2 - n}&= 2^{-7}\\ 2 - n &= -7 \\ \therefore 9 &= n \end{align*}

The ratio between the sum of the first three terms of a geometric series and the sum of the $\text{4}$$^{\text{th}}$, $\text{5}$$^{\text{th}}$ and $\text{6}$$^{\text{th}}$ terms of the same series is $8:27$. Determine the constant ratio and the first $\text{2}$ terms if the third term is $\text{8}$.

\begin{align*} T_{1} + T_{2} + T_{3} &= a + ar + ar^{2} \\ &= a(1 + r + r^{2}) \\ T_{4} + T_{5} + T_{6} &= ar^{3} + ar^{4} + ar^{5} \\ &= ar^{3}(1 + r + r^{2}) \\ \therefore \frac{T_{1} + T_{2} + T_{3} }{T_{4} + T_{5} + T_{6}} &= \frac{a(1 + r + r^{2})}{ar^{3}(1 + r + r^{2})} \\ \text{And } \quad \frac{T_{1} + T_{2} + T_{3} }{T_{4} + T_{5} + T_{6}} &= \frac{8}{27} \\ \therefore \frac{8}{27} &= \frac{a(1 + r + r^{2})}{ar^{3}(1 + r + r^{2})} \\ &=\frac{1}{r^{3}} \\ \therefore r^{3} &= \frac{27}{8} \\ &= \left( \frac{3}{2} \right)^{3} \\ \therefore r &= \frac{3}{2} \end{align*} \begin{align*} \text{And } T_{3} &= 8 \\ \therefore ar^{2} &= 8 \\ a \left( \frac{3}{2} \right)^{2} &= 8 \\ \therefore a &= 8 \times \frac{4}{9} \\ \therefore T_{1} &= \frac{32}{9} \\ T_{2} &= ar \\ &= \frac{32}{9} \times \frac{3}{2} \\ &= \frac{16}{3} \end{align*}

1.4 Finite arithmetic series

Table of Contents

1.6 Infinite series

1.5 Finite geometric series (EMCDZ)

General formula for a finite geometric series (EMCF2)

Worked example 11: Sum of a geometric series

Write down the first three terms of the series

Determine the values of aa and rr

Use the general formula to find the sum of the series

Write the final answer

Worked example 12: Sum of a geometric series

Determine the values of aa and rr

Use the general formula for the sum of a geometric series to determine the value of nn

Write the final answer

Worked example 13: Sum of a geometric series

Write down the first three terms of the series

Determine the values of aa and rr

Calculate the sum of the first eight terms of the geometric series

Write the final answer

Sum of a geometric series

Determine the values of $a$ and $r$

Determine the values of $a$ and $r$

Use the general formula for the sum of a geometric series to determine the value of $n$

Determine the values of $a$ and $r$